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.

kj/mol

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:

1. \( 2\text{H}(g) ightarrow \text{H}_2(g) \quad \Delta H_1^\circ = -436.4 \, \frac{\text{kJ}}{\text{mol}} \)
2. \( 2\text{Br}(g) ightarrow \text{Br}_2(g) \quad \Delta H_2^\circ = -192.5 \, \frac{\text{kJ}}{\text{mol}} \)
3. \( 2\text{HBr}(g) ightarrow \text{H}_2(g) + \text{Br}_2(g) \quad \Delta H_3^\circ = 72.4 \, \frac{\text{kJ}}{\text{mol}} \)

Step2: Manipulate reactions to sum to target

First, reverse reaction 3 and divide by 2, reverse reaction 1 and divide by 2, reverse reaction 2 and divide by 2? Wait, no. Let's think about Hess's law. The target reaction is \( \text{H}(g) + \text{Br}(g) ightarrow \text{HBr}(g) \). Let's see the stoichiometry.

If we take reaction 1: \( 2\text{H}(g) ightarrow \text{H}_2(g) \), divide by 2: \( \text{H}(g) ightarrow \frac{1}{2}\text{H}_2(g) \), \( \Delta H = \frac{\Delta H_1^\circ}{2} = \frac{-436.4}{2} = -218.2 \, \frac{\text{kJ}}{\text{mol}} \)

Reaction 2: \( 2\text{Br}(g) ightarrow \text{Br}_2(g) \), divide by 2: \( \text{Br}(g) ightarrow \frac{1}{2}\text{Br}_2(g) \), \( \Delta H = \frac{\Delta H_2^\circ}{2} = \frac{-192.5}{2} = -96.25 \, \frac{\text{kJ}}{\text{mol}} \)

Reaction 3: \( 2\text{HBr}(g) ightarrow \text{H}_2(g) + \text{Br}_2(g) \), reverse it: \( \text{H}_2(g) + \text{Br}_2(g) ightarrow 2\text{HBr}(g) \), \( \Delta H = -\Delta H_3^\circ = -72.4 \, \frac{\text{kJ}}{\text{mol}} \), then divide by 2: \( \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g) ightarrow \text{HBr}(g) \), \( \Delta H = \frac{-72.4}{2} = -36.2 \, \frac{\text{kJ}}{\text{mol}} \)

Now, add the three manipulated reactions:

1. \( \text{H}(g) ightarrow \frac{1}{2}\text{H}_2(g) \quad \Delta H = -218.2 \)
2. \( \text{Br}(g) ightarrow \frac{1}{2}\text{Br}_2(g) \quad \Delta H = -96.25 \)
3. \( \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g) ightarrow \text{HBr}(g) \quad \Delta H = -36.2 \)

Adding them together: \( \text{H}(g) + \text{Br}(g) + \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g) ightarrow \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g) + \text{HBr}(g) \)

The \( \frac{1}{2}\text{H}_2(g) \) and \( \frac{1}{2}\text{Br}_2(g) \) cancel out, leaving \( \text{H}(g) + \text{Br}(g) ightarrow \text{HBr}(g) \)

Now, sum the \( \Delta H \) values: \( -218.2 - 96.25 - 36.2 = -350.65 \)? Wait, that can't be right. Wait, maybe a better approach. Let's use Hess's law by combining the reactions.

The target reaction is \( \text{H}(g) + \text{Br}(g) ightarrow \text{HBr}(g) \). Let's see the given reactions:

We can write the target reaction as:

\( \frac{1}{2}(2\text{H}(g) ightarrow \text{H}_2(g)) + \frac{1}{2}(2\text{Br}(g) ightarrow \text{Br}_2(g)) + \frac{1}{2}(\text{H}_2(g) + \text{Br}_2(g) ightarrow 2\text{HBr}(g)) \)

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

Then, divide reaction 1 by 2: \( \text{H}(g) ightarrow \frac{1}{2}\text{H}_2(g) \), \( \Delta H = \frac{\Delta H_1^\circ}{2} = -218.2 \)

Divide reaction 2 by 2: \( \text{Br}(g) ightarrow \frac{1}{2}\text{Br}_2(g) \), \( \Delta H = \frac{\Delta H_2^\circ}{2} = -96.25 \)

Divide reversed reaction 3 by 2: \( \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g) ightarrow \text{HBr}(g) \), \( \Delta H = \frac{-\Delta H_3^\circ}{2} = \frac{-72.4}{2} = -36.2 \)

Now, add these three:

\( \text{H}(g) ightarrow \frac{1}{2}\text{H}_2(g) \) (ΔH=-218.2)

\( \text{Br}(g) ightarrow \frac{1}{2}\text{Br}_2(g) \) (ΔH=-96.25)

\( \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g) ightarrow \text{HBr}(g) \) (ΔH=-36.2)

Adding them:

\( \text{H}(g) + \text{Br}(g) + \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g) ightarrow \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g) + \text{HBr}(g) \)

Cancel the common terms:

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

Now, sum the ΔH:

-218.2 + (-96.25) + (-36.2) = -218.2 -96.25 -36.2 = -350.65? Wait, that's not correct. Wait, maybe I messed up the signs. Let's start over.

The target reaction is \( \text{H}(g) + \text{Br}(g) ightarrow \text{HBr}(g) \). Let's express this reaction as a combination of the given reactions.

Given reactions:

1. \( 2\text{H}(g) ightarrow \text{H}_2(g) \quad \Delta H_1 = -436.4 \)
2. \( 2\text{Br}(g) ightarrow \text{Br}_2(g) \quad \Delta H_2 = -192.5 \)
3. \( 2\text{HBr}(g) ightarrow \text{H}_2(g) + \text{Br}_2(g) \quad \Delta H_3 = 72.4 \)

We need to get \( \text{H}(g) + \text{Br}(g) ightarrow \text{HBr}(g) \). Let's rearrange the reactions.

First, reverse reaction 3: \( \text{H}_2(g) + \text{Br}_2(g) ightarrow 2\text{HBr}(g) \quad \Delta H = -\Delta H_3 = -72.4 \)

Now, add reaction 1, reaction 2, and reversed reaction 3:

Reaction 1: \( 2\text{H}(g) ightarrow \text{H}_2(g) \quad \Delta H_1 = -436.4 \)

Reaction 2: \( 2\text{Br}(g) ightarrow \text{Br}_2(g) \quad \Delta H_2 = -192.5 \)

Reversed reaction 3: \( \text{H}_2(g) + \text{Br}_2(g) ightarrow 2\text{HBr}(g) \quad \Delta H = -72.4 \)

Adding these three reactions:

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

The ΔH for this reaction is \( \Delta H_1 + \Delta H_2 + (-\Delta H_3) = -436.4 -192.5 -72.4 = -701.3 \, \frac{\text{kJ}}{\text{mol}} \)

Now, the reaction we got is \( 2\text{H}(g) + 2\text{Br}(g) ightarrow 2\text{HBr}(g) \) with ΔH = -701.3 kJ/mol. To get the target reaction \( \text{H}(g) + \text{Br}(g) ightarrow \text{HBr}(g) \), we divide this reaction by 2. So, divide the ΔH by 2:

\( \Delta H^\circ = \frac{-701.3}{2} = -350.65 \, \frac{\text{kJ}}{\text{mol}} \)? Wait, but let's check the significant digits. The given ΔH values: -436.4 (4 sig figs), -192.5 (4 sig figs), 72.4 (3 sig figs). When we add -436.4 -192.5 = -628.9, then -628.9 -72.4 = -701.3. Then divide by 2: -350.65. But 72.4 has 3 sig figs, so the result should have 3 sig figs? Wait, no, let's check the calculations again.

Wait, -436.4 -192.5 = -628.9; -628.9 -72.4 = -701.3. Then divide by 2: -350.65. But let's check the actual Hess's law application.

The reaction \( 2\text{H}(g) + 2\text{Br}(g) ightarrow 2\text{HBr}(g) \) is the sum of reaction 1, reaction 2, and reversed reaction 3. So:

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

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

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

Adding them: \( 2\text{H}(g) + 2\text{Br}(g) + \text{H}_2(g) + \text{Br}_2(g) ightarrow \text{H}_2(g) + \text{Br}_2(g) + 2\text{HBr}(g) \)

Cancel \( \text{H}_2(g) \) and \( \text{Br}_2(g) \): \( 2\text{H}(g) + 2\text{Br}(g) ightarrow 2\text{HBr}(g) \), which is correct. Then ΔH for this is \( -436.4 + (-192.5) + (-72.4) = -436.4 -192.5 -72.4 = -701.3 \, \text{kJ/mol} \). Then, to get \( \text{H}(g) + \text{Br}(g) ightarrow \text{HBr}(g) \), we divide the reaction by 2, so ΔH is \( \frac{-701.3}{2} = -350.65 \, \text{kJ/mol} \). Rounding to the correct number of significant digits: the given ΔH values: -436.4 (4 sig figs), -192.5 (4 sig figs), 72.4 (3 sig figs). When adding/subtracting, the number of decimal places matters, but here we have -436.4 (1 decimal), -192.5 (1 decimal), -72.4 (1 decimal). So the sum is -701.3 (1 decimal). Then dividing by 2: -350.65, which can be rounded to -351 (if 3 sig figs) or -350.7 (if 4 sig figs). Wait, but let's check the actual calculation again.

Wait, maybe I made a mistake in the sign of reversed reaction 3. Reaction 3 is \( 2\text{HBr}(g) ightarrow \text{H}_2(g) + \text{Br}_2(g) \) with ΔH=72.4. So reversing it gives \( \text{H}_2(g) + \text{Br}_2(g) ightarrow 2\text{HBr}(g) \) with ΔH=-72.4. Then adding reaction 1 (ΔH=-436.4) and reaction 2 (ΔH=-192.5):

Total ΔH = -436.4 + (-192.5) + (-72.4) = -436.4 -192.5 -72.4 = -701.3. Then dividing by 2: -350.65. So the ΔH for \( \text{H}(g) + \text{Br}(g) ightarrow \text{HBr}(g) \) is -350.65 kJ/mol, which can be written as -351 kJ/mol (3 sig figs) or -350.7 kJ/mol (4 sig figs). But let's check the problem statement: the given ΔH values have -436.4 (4 sig figs), -192.5 (4 sig figs), 72.4 (3 sig figs). When we add them, the least number of decimal places is 1 (all have 1 decimal), so the sum is -701.3 (1 decimal). Then dividing by 2, we get -350.65, which can be rounded to -351 (if 3 sig figs) or -350.7 (if 4 sig figs). But let's see the target reaction: the coefficients are 1,1,1, so the ΔH should be calculated as follows.

Wait, another way: use Hess's law by expressing the target reaction as:

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

We can write this as:

\( \frac{1}{2}(2\text{H}(g) ightarrow \text{H}_2(g)) + \frac{1}{2}(2\text{Br}(g) ightarrow \text{Br}_2(g)) + \frac{1}{2}(\text{H}_2(g) + \text{Br}_2(g) ightarrow 2\text{HBr}(g)) \)

So the ΔH is:

\( \frac{\Delta H_1}{2} + \frac{\Delta H_2}{2} + \frac{(-\Delta H_3)}{2} \)

Because \( \text{H}_2(g) + \text{Br}_2(g) ightarrow 2\text{HBr}(g) \) has ΔH = -ΔH_3.

So:

\( \Delta H = \frac{-436.4}{2} + \frac{-192.5}{2} + \frac{-72.4}{2} \)

Calculate each term:

\( \frac{-436.4}{2} = -218.2 \)

\( \frac{-192.5}{2} = -96.25 \)

\( \frac{-72.4}{2} = -36.2 \)

Now, sum them: -218.2 -96.25 -36.2 = -350.65

Yes, that's the same result. So the ΔH° for the reaction is -350.65 kJ/mol, which can be rounded to -351 kJ/mol (

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:

1. \( 2\text{H}(g)

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

1. \( 2\text{Br}(g)

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

1. \( 2\text{HBr}(g)

ightarrow \text{H}_2(g) + \text{Br}_2(g) \quad \Delta H_3^\circ = 72.4 \, \frac{\text{kJ}}{\text{mol}} \)

Step2: Manipulate reactions to sum to target

First, reverse reaction 3 and divide by 2, reverse reaction 1 and divide by 2, reverse reaction 2 and divide by 2? Wait, no. Let's think about Hess's law. The target reaction is \( \text{H}(g) + \text{Br}(g)
ightarrow \text{HBr}(g) \). Let's see the stoichiometry.

If we take reaction 1: \( 2\text{H}(g)
ightarrow \text{H}_2(g) \), divide by 2: \( \text{H}(g)
ightarrow \frac{1}{2}\text{H}_2(g) \), \( \Delta H = \frac{\Delta H_1^\circ}{2} = \frac{-436.4}{2} = -218.2 \, \frac{\text{kJ}}{\text{mol}} \)

Reaction 2: \( 2\text{Br}(g)
ightarrow \text{Br}_2(g) \), divide by 2: \( \text{Br}(g)
ightarrow \frac{1}{2}\text{Br}_2(g) \), \( \Delta H = \frac{\Delta H_2^\circ}{2} = \frac{-192.5}{2} = -96.25 \, \frac{\text{kJ}}{\text{mol}} \)

Reaction 3: \( 2\text{HBr}(g)
ightarrow \text{H}_2(g) + \text{Br}_2(g) \), reverse it: \( \text{H}_2(g) + \text{Br}_2(g)
ightarrow 2\text{HBr}(g) \), \( \Delta H = -\Delta H_3^\circ = -72.4 \, \frac{\text{kJ}}{\text{mol}} \), then divide by 2: \( \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g)
ightarrow \text{HBr}(g) \), \( \Delta H = \frac{-72.4}{2} = -36.2 \, \frac{\text{kJ}}{\text{mol}} \)

Now, add the three manipulated reactions:

1. \( \text{H}(g)

ightarrow \frac{1}{2}\text{H}_2(g) \quad \Delta H = -218.2 \)

1. \( \text{Br}(g)

ightarrow \frac{1}{2}\text{Br}_2(g) \quad \Delta H = -96.25 \)

1. \( \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g)

ightarrow \text{HBr}(g) \quad \Delta H = -36.2 \)

Adding them together: \( \text{H}(g) + \text{Br}(g) + \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g)
ightarrow \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g) + \text{HBr}(g) \)

The \( \frac{1}{2}\text{H}_2(g) \) and \( \frac{1}{2}\text{Br}_2(g) \) cancel out, leaving \( \text{H}(g) + \text{Br}(g)
ightarrow \text{HBr}(g) \)

Now, sum the \( \Delta H \) values: \( -218.2 - 96.25 - 36.2 = -350.65 \)? Wait, that can't be right. Wait, maybe a better approach. Let's use Hess's law by combining the reactions.

The target reaction is \( \text{H}(g) + \text{Br}(g)
ightarrow \text{HBr}(g) \). Let's see the given reactions:

We can write the target reaction as:

\( \frac{1}{2}(2\text{H}(g)
ightarrow \text{H}_2(g)) + \frac{1}{2}(2\text{Br}(g)
ightarrow \text{Br}_2(g)) + \frac{1}{2}(\text{H}_2(g) + \text{Br}_2(g)
ightarrow 2\text{HBr}(g)) \)

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

Then, divide reaction 1 by 2: \( \text{H}(g)
ightarrow \frac{1}{2}\text{H}_2(g) \), \( \Delta H = \frac{\Delta H_1^\circ}{2} = -218.2 \)

Divide reaction 2 by 2: \( \text{Br}(g)
ightarrow \frac{1}{2}\text{Br}_2(g) \), \( \Delta H = \frac{\Delta H_2^\circ}{2} = -96.25 \)

Divide reversed reaction 3 by 2: \( \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g)
ightarrow \text{HBr}(g) \), \( \Delta H = \frac{-\Delta H_3^\circ}{2} = \frac{-72.4}{2} = -36.2 \)

Now, add these three:

\( \text{H}(g)
ightarrow \frac{1}{2}\text{H}_2(g) \) (ΔH=-218.2)

\( \text{Br}(g)
ightarrow \frac{1}{2}\text{Br}_2(g) \) (ΔH=-96.25)

\( \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g)
ightarrow \text{HBr}(g) \) (ΔH=-36.2)

Adding them:

\( \text{H}(g) + \text{Br}(g) + \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g)
ightarrow \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(g) + \text{HBr}(g) \)

Cancel the common terms:

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

Now, sum the ΔH:

-218.2 + (-96.25) + (-36.2) = -218.2 -96.25 -36.2 = -350.65? Wait, that's not correct. Wait, maybe I messed up the signs. Let's start over.

The target reaction is \( \text{H}(g) + \text{Br}(g)
ightarrow \text{HBr}(g) \). Let's express this reaction as a combination of the given reactions.

Given reactions:

1. \( 2\text{H}(g)

ightarrow \text{H}_2(g) \quad \Delta H_1 = -436.4 \)

1. \( 2\text{Br}(g)

ightarrow \text{Br}_2(g) \quad \Delta H_2 = -192.5 \)

1. \( 2\text{HBr}(g)

ightarrow \text{H}_2(g) + \text{Br}_2(g) \quad \Delta H_3 = 72.4 \)

We need to get \( \text{H}(g) + \text{Br}(g)
ightarrow \text{HBr}(g) \). Let's rearrange the reactions.

First, reverse reaction 3: \( \text{H}_2(g) + \text{Br}_2(g)
ightarrow 2\text{HBr}(g) \quad \Delta H = -\Delta H_3 = -72.4 \)

Now, add reaction 1, reaction 2, and reversed reaction 3:

Reaction 1: \( 2\text{H}(g)
ightarrow \text{H}_2(g) \quad \Delta H_1 = -436.4 \)

Reaction 2: \( 2\text{Br}(g)
ightarrow \text{Br}_2(g) \quad \Delta H_2 = -192.5 \)

Reversed reaction 3: \( \text{H}_2(g) + \text{Br}_2(g)
ightarrow 2\text{HBr}(g) \quad \Delta H = -72.4 \)

Adding these three reactions:

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

The ΔH for this reaction is \( \Delta H_1 + \Delta H_2 + (-\Delta H_3) = -436.4 -192.5 -72.4 = -701.3 \, \frac{\text{kJ}}{\text{mol}} \)

Now, the reaction we got is \( 2\text{H}(g) + 2\text{Br}(g)
ightarrow 2\text{HBr}(g) \) with ΔH = -701.3 kJ/mol. To get the target reaction \( \text{H}(g) + \text{Br}(g)
ightarrow \text{HBr}(g) \), we divide this reaction by 2. So, divide the ΔH by 2:

\( \Delta H^\circ = \frac{-701.3}{2} = -350.65 \, \frac{\text{kJ}}{\text{mol}} \)? Wait, but let's check the significant digits. The given ΔH values: -436.4 (4 sig figs), -192.5 (4 sig figs), 72.4 (3 sig figs). When we add -436.4 -192.5 = -628.9, then -628.9 -72.4 = -701.3. Then divide by 2: -350.65. But 72.4 has 3 sig figs, so the result should have 3 sig figs? Wait, no, let's check the calculations again.

Wait, -436.4 -192.5 = -628.9; -628.9 -72.4 = -701.3. Then divide by 2: -350.65. But let's check the actual Hess's law application.

The reaction \( 2\text{H}(g) + 2\text{Br}(g)
ightarrow 2\text{HBr}(g) \) is the sum of reaction 1, reaction 2, and reversed reaction 3. So:

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

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

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

Adding them: \( 2\text{H}(g) + 2\text{Br}(g) + \text{H}_2(g) + \text{Br}_2(g)
ightarrow \text{H}_2(g) + \text{Br}_2(g) + 2\text{HBr}(g) \)

Cancel \( \text{H}_2(g) \) and \( \text{Br}_2(g) \): \( 2\text{H}(g) + 2\text{Br}(g)
ightarrow 2\text{HBr}(g) \), which is correct. Then ΔH for this is \( -436.4 + (-192.5) + (-72.4) = -436.4 -192.5 -72.4 = -701.3 \, \text{kJ/mol} \). Then, to get \( \text{H}(g) + \text{Br}(g)
ightarrow \text{HBr}(g) \), we divide the reaction by 2, so ΔH is \( \frac{-701.3}{2} = -350.65 \, \text{kJ/mol} \). Rounding to the correct number of significant digits: the given ΔH values: -436.4 (4 sig figs), -192.5 (4 sig figs), 72.4 (3 sig figs). When adding/subtracting, the number of decimal places matters, but here we have -436.4 (1 decimal), -192.5 (1 decimal), -72.4 (1 decimal). So the sum is -701.3 (1 decimal). Then dividing by 2: -350.65, which can be rounded to -351 (if 3 sig figs) or -350.7 (if 4 sig figs). Wait, but let's check the actual calculation again.

Wait, maybe I made a mistake in the sign of reversed reaction 3. Reaction 3 is \( 2\text{HBr}(g)
ightarrow \text{H}_2(g) + \text{Br}_2(g) \) with ΔH=72.4. So reversing it gives \( \text{H}_2(g) + \text{Br}_2(g)
ightarrow 2\text{HBr}(g) \) with ΔH=-72.4. Then adding reaction 1 (ΔH=-436.4) and reaction 2 (ΔH=-192.5):

Total ΔH = -436.4 + (-192.5) + (-72.4) = -436.4 -192.5 -72.4 = -701.3. Then dividing by 2: -350.65. So the ΔH for \( \text{H}(g) + \text{Br}(g)
ightarrow \text{HBr}(g) \) is -350.65 kJ/mol, which can be written as -351 kJ/mol (3 sig figs) or -350.7 kJ/mol (4 sig figs). But let's check the problem statement: the given ΔH values have -436.4 (4 sig figs), -192.5 (4 sig figs), 72.4 (3 sig figs). When we add them, the least number of decimal places is 1 (all have 1 decimal), so the sum is -701.3 (1 decimal). Then dividing by 2, we get -350.65, which can be rounded to -351 (if 3 sig figs) or -350.7 (if 4 sig figs). But let's see the target reaction: the coefficients are 1,1,1, so the ΔH should be calculated as follows.

Wait, another way: use Hess's law by expressing the target reaction as:

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

We can write this as:

\( \frac{1}{2}(2\text{H}(g)
ightarrow \text{H}_2(g)) + \frac{1}{2}(2\text{Br}(g)
ightarrow \text{Br}_2(g)) + \frac{1}{2}(\text{H}_2(g) + \text{Br}_2(g)
ightarrow 2\text{HBr}(g)) \)

So the ΔH is:

\( \frac{\Delta H_1}{2} + \frac{\Delta H_2}{2} + \frac{(-\Delta H_3)}{2} \)

Because \( \text{H}_2(g) + \text{Br}_2(g)
ightarrow 2\text{HBr}(g) \) has ΔH = -ΔH_3.

So:

\( \Delta H = \frac{-436.4}{2} + \frac{-192.5}{2} + \frac{-72.4}{2} \)

Calculate each term:

\( \frac{-436.4}{2} = -218.2 \)

\( \frac{-192.5}{2} = -96.25 \)

\( \frac{-72.4}{2} = -36.2 \)

Now, sum them: -218.2 -96.25 -36.2 = -350.65

Yes, that's the same result. So the ΔH° for the reaction is -350.65 kJ/mol, which can be rounded to -351 kJ/mol (