- A$12.75$
- ✓$13.06$
- C$12.01$
- D$10.2$
$E_5 -E_1= -0.54 -(-13.6)$
$= 13.06\, eV$
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$C_{(graphite)} +O_2(g) \rightarrow CO_2(g)\,;$
$\Delta _rH^o=-395.5 \, kJ\,mol^{-1}$
$H_2 (g) + \frac{1}{2} O_2 (g) \rightarrow H_2O(l)\,;$ $\Delta _rH^o =-285.8\, kJ\, mol^{-1}$
$CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)\,;$
$\Delta _rH^o = + 890.3\, kJ\, mol^{-1}$
Based on the above thermochemical equations, the value of, $\Delta _rH^o $ at $298\, K$ for the reaction
$C_{(graphite)} + 2H_2(g) \rightarrow CH_4(g) $ will be for $\Delta_{r} H^{\circ}$ ........... $kJ \,mol^{-1}$
$\mathrm{A} \stackrel{700 \mathrm{K}}{\rightarrow}$ Product
$\mathrm{A}\xrightarrow[\text { catalyst }]{500 \mathrm{K}} $ Product
it was found that $\mathrm{E}_{\mathrm{a}}$ is decreased by $30 \;\mathrm{kJ} / \mathrm{mol}$ in the presence of catalyst.
If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same $):$
(Round off to the Nearest Integer) [Given : $R =8.314 \,J mol ^{-1} K ^{-1}$. Assume,
hydrogen is an ideal gas] [Atomic mass off $Fe$ is $55.85\, u$ ]
$(I)$ $B + AN{O_3} \longrightarrow BN{O_3} + A$
$(II)$ $A + HCl \longrightarrow ACl + \frac{1}{2}\,{H_2}$
$(III)$ $D + ECl \longrightarrow DCl + E$
$(IV)$ $D + HN{O_3} \longrightarrow {H_2}$ gas is not evolved