JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
For the reaction, \(A(g) + B(g) \to C(g) + D(g),\) \(\Delta H^o\) and \(\Delta S^o\) are, respectively, \(-29.8\,\,kJ\, mol^{-1}\) and \(-0.100\,\,kJ\,K^{-1}\) \(mol^{-1}\) at \(298\,K.\) The equilibrium constant for the reaction at \(298\,K\) is
- A \(1.0 \times 10^{-10}\)
- B \(10\)
- C \(1\)
- D \(1.0 \times 10^{10}\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
Given \(\Delta {H^o} = - 29.8\,kJ\,mo{l^{ - 1}}\) \(\Delta {S^o} = - 1.00\,kJ\,{K^{ - 1}}\) From the equation \(\Delta G = \Delta {H^o} - T\Delta {S^o} = - 29.8 - (298 \times - 0.100)\) \( = - 29.8 + 29.8 = 0\) Now, \(\Delta {G^o} = - 2.303\,RT\,\log \,{K_{eq}}\)…
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