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 find \(\Delta H^\circ\) for \(H(g)+Br(g) ightarrow HBr(g)\). Let's denote the given reactions as:
1. \(2H(g) ightarrow H_2(g)\) \(\Delta H_1^\circ=-436.4\ \frac{kJ}{mol}\)
2. \(2Br(g) ightarrow Br_2(g)\) \(\Delta H_2^\circ=-192.5\ \frac{kJ}{mol}\)
3. \(2HBr(g) ightarrow H_2(g)+Br_2(g)\) \(\Delta H_3^\circ = 72.4\ \frac{kJ}{mol}\)

The target reaction is \(H(g)+Br(g) ightarrow HBr(g)\), which is half of the reaction \(2H(g)+2Br(g) ightarrow 2HBr(g)\). So first, we need to combine the given reactions to get \(2H(g)+2Br(g) ightarrow 2HBr(g)\) and then divide by 2.

Step2: Combine the given reactions

First, reverse reaction 3: \(H_2(g)+Br_2(g) ightarrow 2HBr(g)\) \(\Delta H_3'^\circ=-\Delta H_3^\circ=- 72.4\ \frac{kJ}{mol}\) (since reversing a reaction changes the sign of \(\Delta H\)).

Now, add reaction 1, reaction 2, and reversed reaction 3:
Reaction 1: \(2H(g) ightarrow H_2(g)\) \(\Delta H_1^\circ=-436.4\ \frac{kJ}{mol}\)
Reaction 2: \(2Br(g) ightarrow Br_2(g)\) \(\Delta H_2^\circ=-192.5\ \frac{kJ}{mol}\)
Reversed reaction 3: \(H_2(g)+Br_2(g) ightarrow 2HBr(g)\) \(\Delta H_3'^\circ=-72.4\ \frac{kJ}{mol}\)

Adding these three reactions: \(2H(g)+2Br(g)+H_2(g)+Br_2(g) ightarrow H_2(g)+Br_2(g)+2HBr(g)\). The \(H_2(g)\) and \(Br_2(g)\) cancel out on both sides, giving \(2H(g)+2Br(g) ightarrow 2HBr(g)\).

Step3: Calculate \(\Delta H\) for \(2H(g)+2Br(g)

ightarrow 2HBr(g)\)
The \(\Delta H\) for the combined reaction is \(\Delta H^\circ=\Delta H_1^\circ+\Delta H_2^\circ+\Delta H_3'^\circ\)
\[ \begin{align*} \Delta H^\circ&=-436.4+(-192.5)+(-72.4)\\ &=-436.4 - 192.5-72.4\\ &=-701.3\ \frac{kJ}{mol} \end{align*} \]

Step4: Find \(\Delta H\) for target reaction

The target reaction \(H(g)+Br(g) ightarrow HBr(g)\) is half of \(2H(g)+2Br(g) ightarrow 2HBr(g)\). So we divide \(\Delta H^\circ\) by 2:
\[ \Delta H^\circ_{target}=\frac{-701.3}{2}=-350.65\ \frac{kJ}{mol} \] Wait, wait, let's recalculate the sum: \(-436.4-192.5 = -628.9\); \(-628.9-72.4=-701.3\). Then divide by 2: \(\frac{-701.3}{2}=-350.65\)? Wait, no, wait. Wait, the reversed reaction 3 is \(H_2(g)+Br_2(g) ightarrow 2HBr(g)\) with \(\Delta H=-72.4\). Then adding reaction 1 (\(2H ightarrow H_2\), \(\Delta H=-436.4\)) and reaction 2 (\(2Br ightarrow Br_2\), \(\Delta H=-192.5\)):

So total \(\Delta H\) for \(2H + 2Br ightarrow 2HBr\) is \(\Delta H_1+\Delta H_2+\Delta H_{reverse\ 3}=-436.4-192.5 - 72.4\). Let's calculate that: \(-436.4-192.5=-628.9\); \(-628.9 - 72.4=-701.3\). Then divide by 2: \(\frac{-701.3}{2}=-350.65\)? Wait, but that seems off. Wait, maybe I made a mistake in the sign. Wait, the target reaction is \(H + Br ightarrow HBr\). Let's try another approach.

Alternative approach: The target reaction is \(H(g)+Br(g) ightarrow HBr(g)\). Let's find the enthalpy of formation or use Hess's law correctly.

The reaction \(2H(g)+2Br(g) ightarrow 2HBr(g)\) can be obtained by:

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 reversed: \(H_2(g)+Br_2(g) ightarrow 2HBr(g)\) \(\Delta H_3'=-72.4\) (since original reaction 3 is \(2HBr ightarrow H_2 + Br_2\) with \(\Delta H = 72.4\), so reverse is \(-72.4\))

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

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

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

\(\Delta H=\Delta H_1+\Delta H_2+\Delta H_3'=-436.4-192.5 - 72.4=-701.3\ kJ/mol\) for \(2\ moles\) of \(HBr\) formation.

Now, the target reaction is \(H(g)+Br(g) ightarrow HBr(g)\), which is 1 mole of \(HBr\) formation from \(H\) and \(Br\) atoms. So we divide the \(\Delta H\) by 2:

\(\Delta H=\frac{-701.3}{2}=-350.65\ kJ/mol\)? Wait, but that seems incorrect. Wait, maybe I messed up the reactions. Wait, let's check the given reactions again.

Wait, the third reaction is \(2HBr ightarrow H_2 + Br_2\) with \(\Delta H = 72.4\). So the reverse is \(H_2 + Br_2 ightarrow 2HBr\) with \(\Delta H=-72.4\). Then, reaction 1: \(2H ightarrow H_2\) (\(\Delta H=-436.4\)), reaction 2: \(2Br ightarrow Br_2\) (\(\Delta H=-192.5\)). So adding reaction 1, reaction 2, and reverse reaction 3:

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

So \(2H + 2Br ightarrow 2HBr\), \(\Delta H=-436.4-192.5 - 72.4=-701.3\ kJ/mol\). Then, the reaction \(H + Br ightarrow HBr\) is half of that, so \(\Delta H=\frac{-701.3}{2}=-350.65\ kJ/mol\). But let's check significant digits. The given values have three or four significant digits. Wait, maybe I made a mistake in the sign. Wait, no, let's think about bond enthalpies. The formation of \(HBr\) from \(H\) and \(Br\) atoms should be exothermic, so negative \(\Delta H\) makes sense. Wait, but let's recalculate the sum:

\(-436.4 - 192.5 = -628.9\); \(-628.9 - 72.4 = -701.3\). Then divide by 2: \(-350.65\), which rounds to -351 (if we consider significant digits, but let's check the given data. Wait, the first reaction has \(-436.4\) (four sig figs), second \(-192.5\) (four), third \(72.4\) (three). So when we add them: \(-436.4 - 192.5 = -628.9\); \(-628.9 - 72.4 = -701.3\) (four sig figs in the sum? Wait, 72.4 has three, so the sum should have three decimal places? No, significant digits for addition: the number of decimal places is determined by the least precise measurement. \(-436.4\) (one decimal), \(-192.5\) (one decimal), \(72.4\) (one decimal). So the sum \(-436.4 - 192.5 = -628.9\) (one decimal), then \(-628.9 - 72.4 = -701.3\) (one decimal). Then dividing by 2: \(-350.65\), which can be rounded to -351 (three significant digits) or -350.7 (four). Wait, but maybe I made a mistake in the reaction combination.

Wait, another way: Let's write the target reaction as \(\frac{1}{2}(2H(g)+2Br(g) ightarrow 2HBr(g))\). So first, find \(\Delta H\) for \(2H(g)+2Br(g) ightarrow 2HBr(g)\).

From the given reactions:

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. \(H_2(g)+Br_2(g) ightarrow 2HBr(g)\) \(\Delta H_3=-72.4\) (reverse of reaction 3)

Now, add reactions 1, 2, and 3:

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

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

\(\Delta H=\Delta H_1+\Delta H_2+\Delta H_3=-436.4-192.5 - 72.4=-701.3\ kJ/mol\)

Then, the target reaction is \(H(g)+Br(g) ightarrow HBr(g)\), so divide by 2:

\(\Delta H=\frac{-701.3}{2}=-350.65\ \frac{kJ}{mol}\)

Answer:

\(\boxed{-351}\) (or \(-350.7\) depending on rounding, but let's check the calculation again. Wait, maybe I made a mistake. Wait, let's recalculate the sum:

\(-436.4 + (-192.5) + (-72.4) = -436.4 - 192.5 - 72.4 = - (436.4 + 192.5 + 72.4) = - (436.4 + 264.9) = -701.3\). Then divide by 2: \(-350.65\), which is approximately \(-351\) when rounded to three significant digits (since 72.4 has three, so the final answer should have three).