JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
Consider the following reaction approaching equilibrium at \(27^{\circ}\,C\) and \(1\,atm\) pressure The standard Gibb's energy change \(\left(\Delta_{ r } G ^{\circ}\right)\) at \(27^{\circ}\,C\) is \((-)\) \(..........kJ\,mol ^{-1}\) (Nearest integer). (Given : \(R =8.3\,J\,K ^{-1}\, mol ^{-1}\) and \(\ln 10=2.3\) )
- A \(6\)
- B \(3\)
- C \(12\)
- D \(9\)
Answer & Solution
Correct Answer
(A) \(6\)
Step-by-step Solution
Detailed explanation
\(\because \Delta G ^{\circ}=- RT \ln K _{ eq }\) \(\text { and } K _{ eq }=\frac{ K _{ f }}{ K _{ b }}\) \(\therefore K _{ eq }=\frac{10^3}{10^2}=10\) \(\therefore \Delta G =- RT \ln 10\) \(\Rightarrow-(8.3 \times 300 \times 2.3)=-5.7\,kJ\,mole ^{-1} \approx 6\,kJ\)…
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