JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
For the reaction \(A ( g ) \rightleftharpoons B ( g )\) at \(495\, K\) \(\Delta_{ I } G ^{\circ}=-9.478\, kJ\, mol ^{-1}\) If we start the reaction in a closed container at \(495\, K\) with \(22\) millimoles of \(A ,\) the amount of \(B\) is the equilibrium mixture is millimoles .............. (Round off to the Nearest Integer). \(\left[ R =8.314 J mol ^{-1} K ^{-1} ; \ell n 10=2.303\right]\)
- A \(25\)
- B \(30\)
- C \(20\)
- D \(35\)
Answer & Solution
Correct Answer
(C) \(20\)
Step-by-step Solution
Detailed explanation
\(\Delta G ^{\circ}=- RT \ell n K _{ eq }\) Given \(\Delta G ^{\circ}=-9.478 KJ / mole\) So \(-9.478 \times 10^{3}=-495 \times 8.314 \times \ell n K _{ eq }\) \(\ell n K _{ eq }=2.303\) \(=\ell n 10\) So \(K _{ eq }=10\) Now \(A ( g ) \rightleftharpoons B ( g )\)…
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