Question

you are given the following data:
2h(g)→h₂(g) δh°=-436.4 kj/mol
2br(g)→br₂(g) δh°=-192.5 kj/mol
2hbr(g)→h₂(g)+br₂(g) δh°=72.4 kj/mol
calculate δh° for the reaction
h(g)+br(g)→hbr(g)
be sure your answer has the correct number of significant digits.

Explanation:

Step1: Analyze given reactions and target reaction

We have three given reactions and need to manipulate them to get the target reaction \( \text{H}(g) + \text{Br}(g) ightarrow \text{HBr}(g) \). Let's label the given reactions as:

Reaction 1: \( 2\text{H}(g) ightarrow \text{H}_2(g) \), \( \Delta H_1^\circ = -436.4 \frac{\text{kJ}}{\text{mol}} \)

Reaction 2: \( 2\text{Br}(g) ightarrow \text{Br}_2(g) \), \( \Delta H_2^\circ = -192.5 \frac{\text{kJ}}{\text{mol}} \)

Reaction 3: \( 2\text{HBr}(g) ightarrow \text{H}_2(g) + \text{Br}_2(g) \), \( \Delta H_3^\circ = 72.4 \frac{\text{kJ}}{\text{mol}} \)

Step2: Reverse and adjust Reaction 3

Reverse Reaction 3 to get \( \text{H}_2(g) + \text{Br}_2(g) ightarrow 2\text{HBr}(g) \). When we reverse a reaction, the sign of \( \Delta H \) changes. So the new \( \Delta H_3'^\circ = -72.4 \frac{\text{kJ}}{\text{mol}} \)

Step3: Add Reaction 1, Reaction 2, and reversed Reaction 3

Now, add Reaction 1, Reaction 2, and the reversed Reaction 3:

\( 2\text{H}(g) ightarrow \text{H}_2(g) \) (Reaction 1)

\( 2\text{Br}(g) ightarrow \text{Br}_2(g) \) (Reaction 2)

\( \text{H}_2(g) + \text{Br}_2(g) ightarrow 2\text{HBr}(g) \) (Reversed Reaction 3)

Adding these together, the \( \text{H}_2(g) \) and \( \text{Br}_2(g) \) cancel out on both sides, giving:

\( 2\text{H}(g) + 2\text{Br}(g) ightarrow 2\text{HBr}(g) \)

Now, calculate the total \( \Delta H \) for this combined reaction:

\( \Delta H_{\text{total}}^\circ = \Delta H_1^\circ + \Delta H_2^\circ + \Delta H_3'^\circ \)

\( \Delta H_{\text{total}}^\circ = -436.4 + (-192.5) + (-72.4) \)

\( \Delta H_{\text{total}}^\circ = -436.4 - 192.5 - 72.4 = -701.3 \frac{\text{kJ}}{\text{mol}} \)

Step4: Adjust for the target reaction

The combined reaction we got is \( 2\text{H}(g) + 2\text{Br}(g) ightarrow 2\text{HBr}(g) \) with \( \Delta H_{\text{total}}^\circ = -701.3 \frac{\text{kJ}}{\text{mol}} \). But our target reaction is \( \text{H}(g) + \text{Br}(g) ightarrow \text{HBr}(g) \), which is half of the combined reaction. So we divide the \( \Delta H \) by 2:

\( \Delta H^\circ = \frac{\Delta H_{\text{total}}^\circ}{2} = \frac{-701.3}{2} = -350.65 \frac{\text{kJ}}{\text{mol}} \)

Wait, let's check the calculations again. Wait, maybe I made a mistake in the sign when reversing Reaction 3. Let's re - do the calculation:

Reaction 1: \( 2H(g) ightarrow H_2(g)\), \( \Delta H_1=-436.4\)

Reaction 2: \( 2Br(g) ightarrow Br_2(g)\), \( \Delta H_2 = - 192.5\)

Reaction 3: \( 2HBr(g) ightarrow H_2(g)+Br_2(g)\), \( \Delta H_3 = 72.4\). So the reverse of Reaction 3 is \( H_2(g)+Br_2(g) ightarrow 2HBr(g)\), \( \Delta H_3'=-72.4\)

Now, add Reaction 1 + Reaction 2 + Reaction 3':

\( 2H(g)+2Br(g)+H_2(g)+Br_2(g) ightarrow H_2(g)+Br_2(g)+2HBr(g)\)

Simplify: \( 2H(g)+2Br(g) ightarrow 2HBr(g)\)

The total \( \Delta H=\Delta H_1+\Delta H_2+\Delta H_3'=-436.4-192.5 - 72.4=-701.3\)

Now, for the reaction \( H(g)+Br(g) ightarrow HBr(g)\), we can see that it is \( \frac{1}{2}\) of \( 2H(g)+2Br(g) ightarrow 2HBr(g)\)

So \( \Delta H=\frac{-701.3}{2}=-350.65\). But let's check the significant figures. The given values: - 436.4 (4 sig figs), - 192.5 (4 sig figs), 72.4 (3 sig figs). When adding and dividing, the least number of decimal places? Wait, - 436.4 has one decimal place, - 192.5 has one decimal place, - 72.4 has one decimal place. The sum is - 701.3 (one decimal place). Dividing by 2 gives - 350.65, but we should consider significant figures. Wait, maybe another approach.

Alternative approach: Use Hess's law. Let's denote the target reaction as R: \( H(g)+Br(g) ightarrow HBr(g)\)

We can express R in terms of the given reactions.

Given:

1. \( 2H(g) ightarrow H_2(g)\), \( \Delta H_1=-436.4\)

2. \( 2Br(g) ightarrow Br_2(g)\), \( \Delta H_2=-192.5\)

3. \( 2HBr(g) ightarrow H_2(g)+Br_2(g)\), \( \Delta H_3 = 72.4\)

We want to find \( \Delta H \) for \( H(g)+Br(g) ightarrow HBr(g)\)

Let's multiply the target reaction by 2: \( 2H(g)+2Br(g) ightarrow 2HBr(g)\) (let's call this Reaction 4)

Reaction 4 can be obtained by: Reaction 1 + Reaction 2 - Reaction 3 (because Reaction 3 is \( 2HBr ightarrow H_2 + Br_2\), so to get \( H_2+Br_2 ightarrow 2HBr\), we subtract Reaction 3 or reverse it)

So \( \Delta H_4=\Delta H_1+\Delta H_2-\Delta H_3\)

\( \Delta H_4=-436.4-192.5 - 72.4=-701.3\)

Then, since Reaction 4 is 2 times the target reaction, \( \Delta H_{\text{target}}=\frac{\Delta H_4}{2}=\frac{-701.3}{2}=-350.65\approx - 351\)? Wait, no, - 350.65 is - 350.7 when rounded to one decimal place? Wait, - 436.4, - 192.5, 72.4. Let's recalculate \( \Delta H_1+\Delta H_2-\Delta H_3\):

- 436.4 - 192.5 = - 628.9; - 628.9 - 72.4=-701.3. Then divide by 2: - 350.65. Now, check the significant figures. The values: - 436.4 (4 sig figs), - 192.5 (4 sig figs), 72.4 (3 sig figs). When adding and subtracting, the number of decimal places: all have one decimal place, so the sum - 701.3 has one decimal place. Dividing by 2 (exact number, so no change in sig figs), so - 350.65, which can be written as - 350.7 (if we consider one decimal place) or - 351 (if we consider three sig figs? Wait, 72.4 has three sig figs, - 436.4 has four, - 192.5 has four. When adding/subtracting, the result should have the same number of decimal places as the least precise measurement. 72.4 has one decimal place, - 436.4 and - 192.5 also have one decimal place. So the sum - 701.3 has one decimal place. Dividing by 2 (a pure number, so the number of sig figs remains the same in terms of decimal places? Wait, no, dividing by 2 is a multiplication by 0.5, which is exact. So - 701.3 / 2 = - 350.65, which should be rounded to one decimal place? Wait, - 350.65 is - 350.7 when rounded to one decimal place. But let's check the problem statement. The given \( \Delta H \) values: - 436.4 (four sig figs), - 192.5 (four sig figs), 72.4 (three sig figs). When we do \( \Delta H_1+\Delta H_2-\Delta H_3\), the result's precision is determined by the least precise term, which is 72.4 (three sig figs in the value, one decimal place). So the sum is - 701.3 (one decimal place, four sig figs? Wait, - 436.4 is four sig figs, - 192.5 is four, 72.4 is three. When adding, the number of decimal places is determined by the least number of decimal places, which is one (all have one decimal place). So the sum is - 701.3 (one decimal place, four sig figs? No, - 701.3 has five significant figures? Wait, no: - 436.4 (4), - 192.5 (4), 72.4 (3). The sum: - 436.4 - 192.5 = - 628.9; - 628.9 - 72.4=-701.3. - 701.3 has four significant figures? Wait, 701.3: the first non - zero digit is 7, then 0,1,3. So four significant figures. Then dividing by 2 (exact) gives - 350.65, which we can round to - 351 (three significant figures? Wait, no, 72.4 has three significant figures, so the final answer should have three significant figures? Wait, no, let's check the calculation again.

Wait, maybe a better way:

The target reaction is \( H(g)+Br(g) ightarrow HBr(g)\)

Let's express the enthalpy change as follows:

We know that:

\( \Delta H_f(H_2)=0\) (by definition), \( \Delta H_f(Br_2)=0\) (by definition)

For the reaction \( 2H(g) ightarrow H_2(g)\), \( \Delta H_1 = 2\Delta H_f(H_2)-2\Delta H_f(H(g))\), so \( - 436.4=0 - 2\Delta H_f(H(g))\), so \( \Delta H_f(H(g)) = 218.2\)

For the reaction \( 2Br(g) ightarrow Br_2(g)\), \( \Delta H_2=2\Delta H_f(Br_2)-2\Delta H_f(Br(g))\), so \( - 192.5 = 0-2\Delta H_f(Br(g))\), so \( \Delta H_f(Br(g))=96.25\)

For the reaction \( 2HBr(g) ightarrow H_2(g)+Br_2(g)\), \( \Delta H_3=\Delta H_f(H_2)+\Delta H_f(Br_2)-2\Delta H_f(HBr(g))\), so \( 72.4 = 0 + 0-2\Delta H_f(HBr(g))\), so \( \Delta H_f(HBr(g))=-36.2\)

Now, for the reaction \( H(g)+Br(g) ightarrow HBr(g)\), \( \Delta H=\Delta H_f(HBr(g))-\Delta H_f(H(g))-\Delta H_f(Br(g))\)

\( \Delta H=-36.2 - 218.2-96.25=-350.65\)

So the enthalpy change is - 351 kJ/mol (if we take three significant figures) or - 350.7 kJ/mol. But looking at the given data, 72.4 has three significant figures, - 436.4 and - 192.5 have four. When we do the calculation \( (-436.4-192.5 - 72.4)/2=(-701.3)/2=-350.65\). Rounding to three significant figures, it's - 351. But let's check the decimal places. - 436.4 has one decimal place, - 192.5 has one decimal place, 72.4 has one decimal place. The sum is - 701.3 (one decimal place), divided by 2 is - 350.65, which can be written as - 350.7 (one decimal place) or - 351 (three significant figures). However, let's check the original problem's data. The value 72.4 has three significant figures, so the answer should have three significant figures. So - 351? Wait, no, - 350.65 is closer to - 351? Wait, no, - 350.65 rounded to three significant figures is - 351? Wait, 350.65: the first three significant figures are 3,5,0. The next digit is 6, which is more than 5, so we round up the third significant figure: 351.

But wait, let's check with Hess's law again.

Reaction 1: \( 2H(g) ightarrow H_2(g)\), \( \Delta H_1=-436.4\)

Reaction 2: \( 2Br(g) ightarrow Br_2(g)\), \( \Delta H_2=-192.5\)

Reaction 3: \( H_2(g)+Br_2(g) ightarrow 2HBr(g)\), \( \Delta H_3'=-72.4\)

Add Reaction 1 + Reaction 2 + Reaction 3':

\( 2H(g)+2Br(g)+H_2(g)+Br_2(g) ightarrow H_2(g)+Br_2(g)+2HBr(g)\)

Simplify: \( 2H(g)+2Br(g) ightarrow 2HBr(g)\), \( \Delta H=-436.4-192.5 - 72.4=-701.3\)

Divide by 2: \( H(g)+Br(g) ightarrow HBr(g)\), \( \Delta H=\frac{-701.3}{2}=-350.65\approx - 351\) (if we take three significant figures) or - 350.7 (if we take one decimal place). But the given values: - 436.4 (four sig figs), - 192.5 (four sig figs), 72.4 (three sig figs). When adding and subtracting, the result should have the same number of decimal places as the least precise measurement (which is one decimal place for all). Then dividing by 2 (a pure number) doesn't change the number of decimal places. So - 701.3 / 2=-350.65, which can be written as - 350.7 (one decimal place) or - 351 (three significant figures). But let's check the problem statement: "Be sure your answer has the correct number of significant digits." The value 72.4 has three significant digits, - 436.4 and - 192.5 have four. When we do the calculation \( (-436.4 - 192.5-72.4)\), the number of significant digits in the result is determined by the least number of significant digits in the terms? No, when adding/subtracting, it's about decimal places. When multiplying/dividing, it's about significant digits. Here, we first add/subtract, then divide.

The sum \( - 436.4-192.5 - 72.4=-701.3\) (one decimal place, four significant digits). Then dividing by 2 (exact) gives - 350.65, which we can present as - 351

Answer:

Step1: Analyze given reactions and target reaction

We have three given reactions and need to manipulate them to get the target reaction \( \text{H}(g) + \text{Br}(g)
ightarrow \text{HBr}(g) \). Let's label the given reactions as:

Reaction 1: \( 2\text{H}(g)
ightarrow \text{H}_2(g) \), \( \Delta H_1^\circ = -436.4 \frac{\text{kJ}}{\text{mol}} \)

Reaction 2: \( 2\text{Br}(g)
ightarrow \text{Br}_2(g) \), \( \Delta H_2^\circ = -192.5 \frac{\text{kJ}}{\text{mol}} \)

Reaction 3: \( 2\text{HBr}(g)
ightarrow \text{H}_2(g) + \text{Br}_2(g) \), \( \Delta H_3^\circ = 72.4 \frac{\text{kJ}}{\text{mol}} \)

Step2: Reverse and adjust Reaction 3

Reverse Reaction 3 to get \( \text{H}_2(g) + \text{Br}_2(g)
ightarrow 2\text{HBr}(g) \). When we reverse a reaction, the sign of \( \Delta H \) changes. So the new \( \Delta H_3'^\circ = -72.4 \frac{\text{kJ}}{\text{mol}} \)

Step3: Add Reaction 1, Reaction 2, and reversed Reaction 3

Now, add Reaction 1, Reaction 2, and the reversed Reaction 3:

\( 2\text{H}(g)
ightarrow \text{H}_2(g) \) (Reaction 1)

\( 2\text{Br}(g)
ightarrow \text{Br}_2(g) \) (Reaction 2)

\( \text{H}_2(g) + \text{Br}_2(g)
ightarrow 2\text{HBr}(g) \) (Reversed Reaction 3)

Adding these together, the \( \text{H}_2(g) \) and \( \text{Br}_2(g) \) cancel out on both sides, giving:

\( 2\text{H}(g) + 2\text{Br}(g)
ightarrow 2\text{HBr}(g) \)

Now, calculate the total \( \Delta H \) for this combined reaction:

\( \Delta H_{\text{total}}^\circ = \Delta H_1^\circ + \Delta H_2^\circ + \Delta H_3'^\circ \)

\( \Delta H_{\text{total}}^\circ = -436.4 + (-192.5) + (-72.4) \)

\( \Delta H_{\text{total}}^\circ = -436.4 - 192.5 - 72.4 = -701.3 \frac{\text{kJ}}{\text{mol}} \)

Step4: Adjust for the target reaction

The combined reaction we got is \( 2\text{H}(g) + 2\text{Br}(g)
ightarrow 2\text{HBr}(g) \) with \( \Delta H_{\text{total}}^\circ = -701.3 \frac{\text{kJ}}{\text{mol}} \). But our target reaction is \( \text{H}(g) + \text{Br}(g)
ightarrow \text{HBr}(g) \), which is half of the combined reaction. So we divide the \( \Delta H \) by 2:

\( \Delta H^\circ = \frac{\Delta H_{\text{total}}^\circ}{2} = \frac{-701.3}{2} = -350.65 \frac{\text{kJ}}{\text{mol}} \)

Wait, let's check the calculations again. Wait, maybe I made a mistake in the sign when reversing Reaction 3. Let's re - do the calculation:

Reaction 1: \( 2H(g)
ightarrow H_2(g)\), \( \Delta H_1=-436.4\)

Reaction 2: \( 2Br(g)
ightarrow Br_2(g)\), \( \Delta H_2 = - 192.5\)

Reaction 3: \( 2HBr(g)
ightarrow H_2(g)+Br_2(g)\), \( \Delta H_3 = 72.4\). So the reverse of Reaction 3 is \( H_2(g)+Br_2(g)
ightarrow 2HBr(g)\), \( \Delta H_3'=-72.4\)

Now, add Reaction 1 + Reaction 2 + Reaction 3':

\( 2H(g)+2Br(g)+H_2(g)+Br_2(g)
ightarrow H_2(g)+Br_2(g)+2HBr(g)\)

Simplify: \( 2H(g)+2Br(g)
ightarrow 2HBr(g)\)

The total \( \Delta H=\Delta H_1+\Delta H_2+\Delta H_3'=-436.4-192.5 - 72.4=-701.3\)

Now, for the reaction \( H(g)+Br(g)
ightarrow HBr(g)\), we can see that it is \( \frac{1}{2}\) of \( 2H(g)+2Br(g)
ightarrow 2HBr(g)\)

So \( \Delta H=\frac{-701.3}{2}=-350.65\). But let's check the significant figures. The given values: - 436.4 (4 sig figs), - 192.5 (4 sig figs), 72.4 (3 sig figs). When adding and dividing, the least number of decimal places? Wait, - 436.4 has one decimal place, - 192.5 has one decimal place, - 72.4 has one decimal place. The sum is - 701.3 (one decimal place). Dividing by 2 gives - 350.65, but we should consider significant figures. Wait, maybe another approach.

Alternative approach: Use Hess's law. Let's denote the target reaction as R: \( H(g)+Br(g)
ightarrow HBr(g)\)

We can express R in terms of the given reactions.

Given:

1. \( 2H(g)

ightarrow H_2(g)\), \( \Delta H_1=-436.4\)

1. \( 2Br(g)

ightarrow Br_2(g)\), \( \Delta H_2=-192.5\)

1. \( 2HBr(g)

ightarrow H_2(g)+Br_2(g)\), \( \Delta H_3 = 72.4\)

We want to find \( \Delta H \) for \( H(g)+Br(g)
ightarrow HBr(g)\)

Let's multiply the target reaction by 2: \( 2H(g)+2Br(g)
ightarrow 2HBr(g)\) (let's call this Reaction 4)

Reaction 4 can be obtained by: Reaction 1 + Reaction 2 - Reaction 3 (because Reaction 3 is \( 2HBr
ightarrow H_2 + Br_2\), so to get \( H_2+Br_2
ightarrow 2HBr\), we subtract Reaction 3 or reverse it)

So \( \Delta H_4=\Delta H_1+\Delta H_2-\Delta H_3\)

\( \Delta H_4=-436.4-192.5 - 72.4=-701.3\)

Then, since Reaction 4 is 2 times the target reaction, \( \Delta H_{\text{target}}=\frac{\Delta H_4}{2}=\frac{-701.3}{2}=-350.65\approx - 351\)? Wait, no, - 350.65 is - 350.7 when rounded to one decimal place? Wait, - 436.4, - 192.5, 72.4. Let's recalculate \( \Delta H_1+\Delta H_2-\Delta H_3\):

• 436.4 - 192.5 = - 628.9; - 628.9 - 72.4=-701.3. Then divide by 2: - 350.65. Now, check the significant figures. The values: - 436.4 (4 sig figs), - 192.5 (4 sig figs), 72.4 (3 sig figs). When adding and subtracting, the number of decimal places: all have one decimal place, so the sum - 701.3 has one decimal place. Dividing by 2 (exact number, so no change in sig figs), so - 350.65, which can be written as - 350.7 (if we consider one decimal place) or - 351 (if we consider three sig figs? Wait, 72.4 has three sig figs, - 436.4 has four, - 192.5 has four. When adding/subtracting, the result should have the same number of decimal places as the least precise measurement. 72.4 has one decimal place, - 436.4 and - 192.5 also have one decimal place. So the sum - 701.3 has one decimal place. Dividing by 2 (a pure number, so the number of sig figs remains the same in terms of decimal places? Wait, no, dividing by 2 is a multiplication by 0.5, which is exact. So - 701.3 / 2 = - 350.65, which should be rounded to one decimal place? Wait, - 350.65 is - 350.7 when rounded to one decimal place. But let's check the problem statement. The given \( \Delta H \) values: - 436.4 (four sig figs), - 192.5 (four sig figs), 72.4 (three sig figs). When we do \( \Delta H_1+\Delta H_2-\Delta H_3\), the result's precision is determined by the least precise term, which is 72.4 (three sig figs in the value, one decimal place). So the sum is - 701.3 (one decimal place, four sig figs? Wait, - 436.4 is four sig figs, - 192.5 is four, 72.4 is three. When adding, the number of decimal places is determined by the least number of decimal places, which is one (all have one decimal place). So the sum is - 701.3 (one decimal place, four sig figs? No, - 701.3 has five significant figures? Wait, no: - 436.4 (4), - 192.5 (4), 72.4 (3). The sum: - 436.4 - 192.5 = - 628.9; - 628.9 - 72.4=-701.3. - 701.3 has four significant figures? Wait, 701.3: the first non - zero digit is 7, then 0,1,3. So four significant figures. Then dividing by 2 (exact) gives - 350.65, which we can round to - 351 (three significant figures? Wait, no, 72.4 has three significant figures, so the final answer should have three significant figures? Wait, no, let's check the calculation again.

Wait, maybe a better way:

The target reaction is \( H(g)+Br(g)
ightarrow HBr(g)\)

Let's express the enthalpy change as follows:

We know that:

\( \Delta H_f(H_2)=0\) (by definition), \( \Delta H_f(Br_2)=0\) (by definition)

For the reaction \( 2H(g)
ightarrow H_2(g)\), \( \Delta H_1 = 2\Delta H_f(H_2)-2\Delta H_f(H(g))\), so \( - 436.4=0 - 2\Delta H_f(H(g))\), so \( \Delta H_f(H(g)) = 218.2\)

For the reaction \( 2Br(g)
ightarrow Br_2(g)\), \( \Delta H_2=2\Delta H_f(Br_2)-2\Delta H_f(Br(g))\), so \( - 192.5 = 0-2\Delta H_f(Br(g))\), so \( \Delta H_f(Br(g))=96.25\)

For the reaction \( 2HBr(g)
ightarrow H_2(g)+Br_2(g)\), \( \Delta H_3=\Delta H_f(H_2)+\Delta H_f(Br_2)-2\Delta H_f(HBr(g))\), so \( 72.4 = 0 + 0-2\Delta H_f(HBr(g))\), so \( \Delta H_f(HBr(g))=-36.2\)

Now, for the reaction \( H(g)+Br(g)
ightarrow HBr(g)\), \( \Delta H=\Delta H_f(HBr(g))-\Delta H_f(H(g))-\Delta H_f(Br(g))\)

\( \Delta H=-36.2 - 218.2-96.25=-350.65\)

So the enthalpy change is - 351 kJ/mol (if we take three significant figures) or - 350.7 kJ/mol. But looking at the given data, 72.4 has three significant figures, - 436.4 and - 192.5 have four. When we do the calculation \( (-436.4-192.5 - 72.4)/2=(-701.3)/2=-350.65\). Rounding to three significant figures, it's - 351. But let's check the decimal places. - 436.4 has one decimal place, - 192.5 has one decimal place, 72.4 has one decimal place. The sum is - 701.3 (one decimal place), divided by 2 is - 350.65, which can be written as - 350.7 (one decimal place) or - 351 (three significant figures). However, let's check the original problem's data. The value 72.4 has three significant figures, so the answer should have three significant figures. So - 351? Wait, no, - 350.65 is closer to - 351? Wait, no, - 350.65 rounded to three significant figures is - 351? Wait, 350.65: the first three significant figures are 3,5,0. The next digit is 6, which is more than 5, so we round up the third significant figure: 351.

But wait, let's check with Hess's law again.

Reaction 1: \( 2H(g)
ightarrow H_2(g)\), \( \Delta H_1=-436.4\)

Reaction 2: \( 2Br(g)
ightarrow Br_2(g)\), \( \Delta H_2=-192.5\)

Reaction 3: \( H_2(g)+Br_2(g)
ightarrow 2HBr(g)\), \( \Delta H_3'=-72.4\)

Add Reaction 1 + Reaction 2 + Reaction 3':

\( 2H(g)+2Br(g)+H_2(g)+Br_2(g)
ightarrow H_2(g)+Br_2(g)+2HBr(g)\)

Simplify: \( 2H(g)+2Br(g)
ightarrow 2HBr(g)\), \( \Delta H=-436.4-192.5 - 72.4=-701.3\)

Divide by 2: \( H(g)+Br(g)
ightarrow HBr(g)\), \( \Delta H=\frac{-701.3}{2}=-350.65\approx - 351\) (if we take three significant figures) or - 350.7 (if we take one decimal place). But the given values: - 436.4 (four sig figs), - 192.5 (four sig figs), 72.4 (three sig figs). When adding and subtracting, the result should have the same number of decimal places as the least precise measurement (which is one decimal place for all). Then dividing by 2 (a pure number) doesn't change the number of decimal places. So - 701.3 / 2=-350.65, which can be written as - 350.7 (one decimal place) or - 351 (three significant figures). But let's check the problem statement: "Be sure your answer has the correct number of significant digits." The value 72.4 has three significant digits, - 436.4 and - 192.5 have four. When we do the calculation \( (-436.4 - 192.5-72.4)\), the number of significant digits in the result is determined by the least number of significant digits in the terms? No, when adding/subtracting, it's about decimal places. When multiplying/dividing, it's about significant digits. Here, we first add/subtract, then divide.

The sum \( - 436.4-192.5 - 72.4=-701.3\) (one decimal place, four significant digits). Then dividing by 2 (exact) gives - 350.65, which we can present as - 351