Question

question 1: use the following reactions (i - iii) to calculate the standard heat of formation of $c_{3}h_{8}(g)$. (i) $c_{3}h_{8}(g)+5o_{2}(g)\to3co_{2}(g)+4h_{2}o(l)$ $delta h^{circ}=-2302kj$ (ii) $c(graphite)+o_{2}(g)\to co_{2}(g)$ $delta h^{circ}=-394kj$ (iii) $2h_{2}(g)+o_{2}(g)\to2h_{2}o(l)$ $delta h^{circ}=-572kj$ a. - 126 kj b. 24 kj c. - 24 kj d. 126 kj e. 1254 kj

Explanation:

Step1: Reverse reaction (i)

$$\mathrm{3CO_2(g)+4H_2O(l)\to C_3H_8(g)+5O_2(g)}\quad\Delta H^{\circ}= + 2302\ \text{kJ}$$

Step2: Multiply reaction (ii) by 3

$$\mathrm{3C(graphite)+3O_2(g)\to 3CO_2(g)}\quad\Delta H^{\circ}=3\times(- 394)\ \text{kJ}=-1182\ \text{kJ}$$

Step3: Multiply reaction (iii) by 2

$$\mathrm{2H_2(g)+O_2(g)\to 2H_2O(l)}\quad\Delta H^{\circ}=2\times(- 286)\ \text{kJ}=-572\ \text{kJ}$$

Step4: Add the three modified reactions

$$\mathrm{3C(graphite)+4H_2(g)\to C_3H_8(g)}$$
$$\Delta H^{\circ}=(2302 - 1182-572)\ \text{kJ}=-118\ \text{kJ}$$
Wait, there is a mistake. Let's re - calculate:
Reaction (i): $\mathrm{C_3H_8(g)+5O_2(g)\to 3CO_2(g)+4H_2O(l)}\quad\Delta H^{\circ}=-2302\ \text{kJ}$ (reverse it: $\mathrm{3CO_2(g)+4H_2O(l)\to C_3H_8(g)+5O_2(g)}\quad\Delta H^{\circ}= + 2302\ \text{kJ}$)
Reaction (ii): $\mathrm{C(graphite)+O_2(g)\to CO_2(g)}\quad\Delta H^{\circ}=-394\ \text{kJ}$ (multiply by 3: $\mathrm{3C(graphite)+3O_2(g)\to 3CO_2(g)}\quad\Delta H^{\circ}=-1182\ \text{kJ}$)
Reaction (iii): $\mathrm{2H_2(g)+O_2(g)\to 2H_2O(l)}\quad\Delta H^{\circ}=-572\ \text{kJ}$ (divide by 2 and multiply by 4: $\mathrm{4H_2(g)+2O_2(g)\to 4H_2O(l)}\quad\Delta H^{\circ}=-1144\ \text{kJ}$)

Add them:
$$(2302-1182 - 1144)\ \text{kJ}=-24\ \text{kJ}$$

Answer:

C. - 24 kJ