For the reversible reaction,$A_{(s)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$,$\Delta G^{\circ} = -350 \ kJ$,which one of the following statements is true?
- AThe reaction is thermodynamically non-feasible
- BThe entropy change is negative
- CEquilibrium constant is greater than one
- DThe reaction should be instantaneous
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For a hypothetical reversible reaction $\frac{1}{2} A_{2(g)} + \frac{3}{2} B_{2(g)} \rightarrow AB_{3(g)}$; $\Delta H = -20 \, kJ$. If the standard entropies of $A_2, B_2$ and $AB_3$ are $60, 40$ and $50 \, J K^{-1} mol^{-1}$ respectively,at what temperature $(K)$ will the reaction be in equilibrium?
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View SolutionFor the reaction $A \rightleftharpoons B$,find the value of $log_{10}K$. Given: $\Delta_rH^o_{298\,K} = -54.07\, kJ\, mol^{-1}$,$\Delta_rS^o_{298\,K} = 10\, J\, K^{-1}\, mol^{-1}$,$R = 8.314\, J\, K^{-1}\, mol^{-1}$,$2.303 \times 8.314 \times 298 = 5705$.
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View Solution$A$ reaction attains equilibrium when the free energy change accompanying it is
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View SolutionFor a certain reaction at $300 \ K$,$K=10$,then $\Delta G^{\circ}$ for the same reaction is . . . . . . $\times 10^{-1} \ kJ \ mol^{-1}$. (Given $R=8.314 \ J \ K^{-1} \ mol^{-1}$)
DifficultJEE MAIN 2024
View SolutionFor the chemical reaction $X \rightleftharpoons Y$,the standard reaction Gibbs energy depends on temperature $T$ (in $K$) as
${\Delta_r}{G^o}$ (in $kJ \ mol^{-1}$) $= 120 - \frac{3}{8} \ T$
The major component of the reaction mixture at $T$ is
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The major component of the reaction mixture at $T$ is
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