Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution
4. Answer the following questions.
xvi. Calculate the standard enthalpy of formation of CH3OH(l) from the following data
i.CH3OH(l)+ 32 O2(g)→ CO2(g)+ 2H2O(l), ΔH0 = -726 kJ mol-1
ii. C (Graphite) + O2(g)→ CO2(g), ΔcH0 = -393 kJ mol-1
iii. H2(g) + 12 O2(g)→ H2O(l), ΔfH0 = -286 kJ mol-1
Answer:-
**Given:**
Given equations are:
(i) \( \text{CH}_3\text{OH}_{(l)} + \frac{3}{2}\text{O}_2\text{(_g)} \rightarrow \text{CO}_2\text{(_g)} + 2\text{H}_2\text{O}_{(l)}; \Delta_r H^\circ = -726 \text{ kJ mol}^{-1}$
(ii) \( \text{C}_{(\text{graphite})} + \text{O}_2\text{(_g)} \rightarrow \text{CO}_2\text{(_g)}; \Delta_r H^\circ = -393 \text{ kJ mol}^{-1}$
(iii) \( \text{H}_2\text{(_g)} + \frac{1}{2}\text{O}_2\text{(_g)} \rightarrow \text{H}_2\text{O}_{(l)}; \Delta_r H^\circ = -286 \text{ kJ mol}^{-1}$
**To find:**
The standard enthalpy of formation (\( \Delta_f H^\circ \)) of CH3OH(l)
**Calculation:**
The required equation is \( \text{C}_{(\text{graphite})} + 2\text{H}_{2(\text{g})} + \frac{1}{2}\text{O}_{2(\text{g})} \rightarrow \text{CH}_3\text{OH}_{(l)} \)
1. Multiply equation (iii) by 2 and add to equation (ii):
\[
2\text{H}_{2(\text{g})} + \text{O}_{2(\text{g})} \rightarrow 2\text{H}_2\text{O}_{(l)}, \Delta_r H^\circ = -575 \text{ kJ mol}^{-1}
\]
\[
\text{C}_{(\text{graphite})} + \text{O}_{2(\text{g})} \rightarrow \text{CO}_2\text{(_g)}, \Delta_r H^\circ = -393 \text{ kJ mol}^{-1}
\]
\[
\begin{array}{l}
\text{C}_{(\text{graphite})} + 2\text{H}_{2(\text{g})} + 2\text{O}_{2(\text{g})} \rightarrow \text{CO}_2\text{(_g)} + 2\text{H}_2\text{O}_{(l)} \\
\Delta_r H^\circ = -572 - 393 = -965 \text{ kJ mol}^{-1}
\end{array}
\]
2. Reverse equation (i) and add to equation (iv):
\[
\text{CO}_2\text{(_g)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{CH}_3\text{OH}_{(l)} + \frac{3}{2}\text{O}_{2(\text{g})}, \Delta_r H^\circ = 726 \text{ kJ mol}^{-1}
\]
\[
\text{C}_{(\text{graphite})} + 2\text{H}_{2(\text{g})} + \frac{1}{2}2\text{O}_{2(\text{g})} \rightarrow \text{CH}_3\text{OH}_{(l)}
\]
\[
\Delta_f H^\circ = \Delta_r H^\circ = 726 - 965 = -239 \text{ kJ mol}^{-1}
\]
The standard enthalpy of formation (\( \Delta_f H^\circ \)) of CH3OH(l) from the given data is -239 kJ mol-1.
Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution