Given,
\(\mathrm{C_{(graphite)} + O_2 \rightarrow CO_2(g)}\); \(\Delta_r H^\circ = -393.5\ \mathrm{kJ\ mol^{-1}}\)
\(\mathrm{H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)}\); \(\Delta_r H^\circ = -285.8\ \mathrm{kJ\ mol^{-1}}\)
\(\mathrm{CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)}\); \(\Delta_r H^\circ = +890.3\ \mathrm{kJ\ mol^{-1}}\)
Based on the above thermochemical equations, the value of \(\Delta_r H^\circ\) at 298 K for the reaction \(\mathrm{C_{(graphite)} + 2H_2(g) \rightarrow CH_4(g)}\) will be :
\(\mathrm{C_{(graphite)} + O_2 \rightarrow CO_2(g)}\); \(\Delta_r H^\circ = -393.5\ \mathrm{kJ\ mol^{-1}}\)
\(\mathrm{H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)}\); \(\Delta_r H^\circ = -285.8\ \mathrm{kJ\ mol^{-1}}\)
\(\mathrm{CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)}\); \(\Delta_r H^\circ = +890.3\ \mathrm{kJ\ mol^{-1}}\)
Based on the above thermochemical equations, the value of \(\Delta_r H^\circ\) at 298 K for the reaction \(\mathrm{C_{(graphite)} + 2H_2(g) \rightarrow CH_4(g)}\) will be :
Show Hint
- \(+144.0\ \mathrm{kJ\ mol^{-1}}\)
- \(-74.8\ \mathrm{kJ\ mol^{-1}}\)
- \(-144.0\ \mathrm{kJ\ mol^{-1}}\)
- \(+74.8\ \mathrm{kJ\ mol^{-1}}\)
The Correct Option is B
Solution and Explanation
Eq. 1: \(\mathrm{C + O_2 \rightarrow CO_2}\), \(\Delta H_1 = -393.5\).
2 x Eq. 2: \(\mathrm{2H_2 + O_2 \rightarrow 2H_2O}\), \(\Delta H = 2(-285.8) = -571.6\).
Eq. 3: \(\mathrm{CO_2 + 2H_2O \rightarrow CH_4 + 2O_2}\), \(\Delta H_3 = +890.3\).
Adding the three and cancelling \(\mathrm{CO_2}\), \(\mathrm{2H_2O}\) and \(\mathrm{2O_2}\) gives \(\mathrm{C + 2H_2 \rightarrow CH_4}\).
\(\Delta_r H^\circ = -393.5 - 571.6 + 890.3 = -74.8\ \mathrm{kJ\ mol^{-1}}\).
Answer: (2) \(-74.8\ \mathrm{kJ\ mol^{-1}}\)
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