Consider the following reaction :
$2A (g) + B (g) \rightarrow 2D(g)$
$\Delta U^{\circ} = -10 \text{ kJ mol}^{-1}$ and $\Delta S^{\circ} = -44 \text{ J K}^{-1} \text{ mol}^{-1}$ at $298 \text{ K}$.
Identify the correct option with $\Delta G^{\circ}$ for the reaction and spontaneity of the reaction at $298 \text{ K}$.
(Given : $R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1}$)
$2A (g) + B (g) \rightarrow 2D(g)$
$\Delta U^{\circ} = -10 \text{ kJ mol}^{-1}$ and $\Delta S^{\circ} = -44 \text{ J K}^{-1} \text{ mol}^{-1}$ at $298 \text{ K}$.
Identify the correct option with $\Delta G^{\circ}$ for the reaction and spontaneity of the reaction at $298 \text{ K}$.
(Given : $R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1}$)
- A$+ 0.636 \text{ kJ mol}^{-1}$,non-spontaneous
- B$- 0.636 \text{ kJ mol}^{-1}$,spontaneous
- C$- 1.635 \text{ kJ mol}^{-1}$,spontaneous
- D$+ 1.635 \text{ kJ mol}^{-1}$,non-spontaneous
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View SolutionMatch List-$I$ with List-$II$.
List-$I$ $(\text{Partial Derivatives})$ List-$II$ $(\text{Thermodynamic Quantity})$ $(A). \left(\frac{\partial G}{\partial T}\right)_{P}$ $(I). C_P$ $(B). \left(\frac{\partial H}{\partial T}\right)_{P}$ $(II). -S$ $(C). \left(\frac{\partial G}{\partial P}\right)_{T}$ $(III). C_V$ $(D). \left(\frac{\partial U}{\partial T}\right)_{V}$ $(IV). V$
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| $(A). \left(\frac{\partial G}{\partial T}\right)_{P}$ | $(I). C_P$ |
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DifficultJEE MAIN 2025
View SolutionGiven
$C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)};$
$\Delta_rH^o = -393.5 \, kJ \, mol^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)};$
$\Delta_rH^o = -285.8 \, kJ \, mol^{-1}$
$CO_{2(g)} + 2H_2O_{(l)} \rightarrow CH_{4(g)} + 2O_{2(g)};$
$\Delta_rH^o = + 890.3 \, kJ \, mol^{-1}$
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 ........... $kJ \, mol^{-1}$.
$C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)};$
$\Delta_rH^o = -393.5 \, kJ \, mol^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)};$
$\Delta_rH^o = -285.8 \, kJ \, mol^{-1}$
$CO_{2(g)} + 2H_2O_{(l)} \rightarrow CH_{4(g)} + 2O_{2(g)};$
$\Delta_rH^o = + 890.3 \, kJ \, mol^{-1}$
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 ........... $kJ \, mol^{-1}$.
DifficultJEE MAIN 2017
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