AP EAMCET · Chemistry · Thermodynamics (C)
From the following data
\[
\begin{aligned}
& \mathrm{CH}_3 \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \\
& \Delta_r H^{\circ}=-726 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(l) \text {; } \\
& \Delta_r \mathrm{H}^{\circ}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{C}(\text { graphite })+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \text {; } \\
& \Delta_r \mathrm{H}^{\circ}=-393 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
&
\end{aligned}
\]
The standard enthalpy of formation of \(\mathrm{CH}_3 \mathrm{OH}(l)\) in \(\mathrm{kJ} \mathrm{mol}^{-1}\) is
- A -239
- B 239
- C 547
- D -905
Answer & Solution
Correct Answer
(A) -239
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { From Eq. (ii) }+ \text { (iii) - Eq. (i) } \\ & \begin{aligned}-286 \times 2-393-(-726)=7 & =-572-393+726 \\ & =-965+726 \\ & =-239 \mathrm{~kJ} / \mathrm{mol}\end{aligned}\end{aligned}\)
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