Question

given the following thermochemical data, the $delta_{r}h^{circ}$ for the reaction $c_{2}h_{2}(g)+2h_{2}(g)
ightleftharpoons c_{2}h_{6}(g)$ is: 1. $c_{2}h_{2}(g)+\frac{5}{2}o_{2}(g)longrightarrow 2co_{2}(g)+h_{2}o(l)$ $delta_{r}h=-1.30\times 10^{3}kj$ 2. $c_{2}h_{6}(g)+\frac{7}{2}o_{2}(g)longrightarrow 2co_{2}(g)+3h_{2}o(l)$ $delta_{r}h=-1.56\times 10^{3}kj$ 3. $h_{2}(g)+\frac{1}{2}o_{2}(g)longrightarrow h_{2}o(l)$ $delta_{r}h=-2.86\times 10^{2}kj$ select one: a. $-5.46\times 10^{3}kj$ b. $-2.60\times 10^{1}kj$ c. $-3.12\times 10^{2}kj$ d. $+2.60\times 10^{1}kj$

Explanation:

Step1: Identify the target reaction and given reactions

The target reaction is $C_2H_2(g)+2H_2(g)
ightleftharpoons C_2H_6(g)$. The given reactions are:
Reaction 1: $C_2H_2(g)+\frac{5}{2}O_2(g)\longrightarrow 2CO_2(g)+H_2O(l)\quad\Delta H_1 = - 1.30\times10^{3}\text{ kJ}$
Reaction 2: $C_2H_6(g)+\frac{7}{2}O_2(g)\longrightarrow 2CO_2(g)+3H_2O(l)\quad\Delta H_2=- 1.56\times10^{3}\text{ kJ}$
Reaction 3: $H_2(g)+\frac{1}{2}O_2(g)\longrightarrow H_2O(l)\quad\Delta H_3=-286\text{ kJ}$
We need to manipulate these reactions to get the target - reaction.

Step2: Reverse Reaction 2

When we reverse Reaction 2, we get $2CO_2(g)+3H_2O(l)\longrightarrow C_2H_6(g)+\frac{7}{2}O_2(g)\quad\Delta H_{2}^{'}=- \Delta H_2 = 1.56\times10^{3}\text{ kJ}$

Step3: Multiply Reaction 3 by 2

$2H_2(g)+O_2(g)\longrightarrow 2H_2O(l)\quad\Delta H_{3}^{'}=2\times\Delta H_3 = 2\times(- 286\text{ kJ})=-572\text{ kJ}$

Step4: Add the manipulated reactions

Adding Reaction 1, the reversed Reaction 2 and the multiplied Reaction 3:
$(C_2H_2(g)+\frac{5}{2}O_2(g)\longrightarrow 2CO_2(g)+H_2O(l))+(2CO_2(g)+3H_2O(l)\longrightarrow C_2H_6(g)+\frac{7}{2}O_2(g))+(2H_2(g)+O_2(g)\longrightarrow 2H_2O(l))$
The $O_2$, $CO_2$ and $H_2O$ terms cancel out on the left - hand and right - hand sides, and we get $C_2H_2(g)+2H_2(g)
ightleftharpoons C_2H_6(g)$
The $\Delta H$ for the target reaction $\Delta H=\Delta H_1+\Delta H_{2}^{'}+\Delta H_{3}^{'}$
$\Delta H=-1.30\times10^{3}\text{ kJ}+1.56\times10^{3}\text{ kJ}-572\text{ kJ}$
$\Delta H=-1.30\times10^{3}+1560 - 572\text{ kJ}$
$\Delta H=-312\text{ kJ}=-3.12\times10^{2}\text{ kJ}$

Answer:

C. $-3.12\times10^{2}\text{ kJ}$