MCQ
Given
$
\begin{array}{l}
C_{\text {graphite }}+O_2 \rightarrow CO_2(g): \\
\Delta_{r} H^0=-393.5 KJ / mol
\end{array}
$
$
H_2(g)+\frac{1}{2} O_2(g) \rightarrow H_2 O(l)
$
$
\Delta_r H^0=-285.8 kJ / mol
$
$
CO_2(g)+2 H_2 O(l) \rightarrow CH_4(g)+2 O_2(g):
$
$
\Delta_r H^0=+890.3 kJ mol^{-1}
$
Based on the above thermochemical equations, the value of $\Delta_r H^0$ at 298 K for the reaction
$
C_{\text {(graphite) }}+2 H_2(g) \rightarrow CH_4(g) \text { will be : }
$
$
\begin{array}{l}
C_{\text {graphite }}+O_2 \rightarrow CO_2(g): \\
\Delta_{r} H^0=-393.5 KJ / mol
\end{array}
$
$
H_2(g)+\frac{1}{2} O_2(g) \rightarrow H_2 O(l)
$
$
\Delta_r H^0=-285.8 kJ / mol
$
$
CO_2(g)+2 H_2 O(l) \rightarrow CH_4(g)+2 O_2(g):
$
$
\Delta_r H^0=+890.3 kJ mol^{-1}
$
Based on the above thermochemical equations, the value of $\Delta_r H^0$ at 298 K for the reaction
$
C_{\text {(graphite) }}+2 H_2(g) \rightarrow CH_4(g) \text { will be : }
$
- A$+144.0 kJ mol ^{-1}$
- ✓$-74.8 kJ mol ^{-1}$
- C$-144.0 kJ mol ^{-1}$
- D$+74.8 kJ mol ^{-1}$
