Question

1. use hesss law to calculate the enthalpy change, \\( \delta h \\), for the reaction below:

\\( c(\text { graphite })+2 \mathrm{h}_{2}(g)+\frac{1}{2} \mathrm{o}_{2}(g) \
ightarrow \mathrm{ch}_{3} \mathrm{oh}(l) \\)
given the following data:
\\( \mathrm{ch}_{3} \mathrm{oh}(l)+\frac{3}{2} \mathrm{o}_{2}(g) \
ightarrow \mathrm{co}_{2}(g)+2 \mathrm{h}_{2} \mathrm{o}(l) \quad \delta \mathrm{h}^{\circ}=-726.4 \mathrm{~kj} / \mathrm{mol} \\)
\\( \mathrm{c}(\text { graphite })+\mathrm{o}_{2}(g) \
ightarrow \mathrm{co}_{2}(g) \quad \delta \mathrm{h}^{\circ}=-393.5 \mathrm{~kj} / \mathrm{mol} \\)
\\( \mathrm{h}_{2}(g)+\frac{1}{2} \mathrm{o}_{2}(g) \
ightarrow \mathrm{h}_{2} \mathrm{o}(l) \quad \delta \mathrm{h}^{\circ}=-285.8 \mathrm{~kj} / \mathrm{mol} \\)
a. \\( -238.7 \mathrm{~kj} / \mathrm{mol} \\)
b. \\( -47.1 \mathrm{~kj} / \mathrm{mol} \\)
c. \\( +47.1 \mathrm{~kj} / \mathrm{mol} \\)
d. \\( -2811.4 \mathrm{~kj} / \mathrm{mol} \\)
e. \\( +334 \mathrm{~kj} / \mathrm{mol} \\)

Explanation:

Step1: Reverse the first reaction

$$\ce{CO2(g) + 2H2O(l)\to CH3OH(l) + \frac{3}{2}O2(g)}$$
$$\Delta H^{\circ}=+ 726.4\ \text{kJ/mol}$$

Step2: Use the second reaction as it is

$$\ce{C(graphite) + O2(g)\to CO2(g)}$$
$$\Delta H^{\circ}=-393.5\ \text{kJ/mol}$$

Step3: Multiply the third reaction by 2

$$\ce{2H2(g) + O2(g)\to 2H2O(l)}$$
$$\Delta H^{\circ}=2\times(- 285.8)\ \text{kJ/mol}=-571.6\ \text{kJ/mol}$$

Step4: Add the three reactions

$$\ce{CO2(g) + 2H2O(l)+C(graphite) + O2(g)+2H2(g) + O2(g)\to CH3OH(l) + \frac{3}{2}O2(g)+CO2(g)+2H2O(l)}$$
Cancel out $\ce{CO2(g)}$ and $\ce{2H2O(l)}$ on both sides. Also, simplify the oxygen: $O_2 + O_2-\frac{3}{2}O_2=\frac{1}{2}O_2$. We get $$\ce{C(graphite) + 2H2(g)+\frac{1}{2}O2(g)\to CH3OH(l)}$$

Step5: Calculate $\Delta H$

$$\Delta H=(+726.4-393.5 - 571.6)\ \text{kJ/mol}$$
$$\Delta H=-238.7\ \text{kJ/mol}$$

Answer:

A. -238.7 kJ/mol