JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
The gas phase reaction \(2 A ( g ) \rightleftharpoons A _{2}( g )\) at \(400\, K\) has \(\Delta G ^{\circ}=+25.2\, kJ mol ^{-1}\). The equilibrium constant \(K _{ C }\) for this reaction is \(...... \times 10^{-2}\). (Round off to the Nearest integ \(\left[\right.\) Use \(: R=8.3\, J mol ^{-1} K ^{-1}, \ln 10=2.3\) \(\left.\log _{10} 2=0.30,1\, atm =1\, bar \right]\) \([\) antilog \((-0.3)=0.501]\)
- A \(141\)
- B \(166\)
- C \(206\)
- D \(111\)
Answer & Solution
Correct Answer
(B) \(166\)
Step-by-step Solution
Detailed explanation
Using formula \(\Delta_{ r } G ^{0}=- RT \ln K _{ p }\) \(25200=-2.3 \times 8.3 \times 400 \log \left( K _{ p }\right)\) \(K _{ p }=10^{-3.3}=10^{-3} \times 0.501\) \(=5.01 \times 10^{-4} \,Bar ^{-1}\) \(=5.01 \times 10^{-9}\, Pa ^{-1}\) \(=\frac{ K _{ C }}{8.3 \times 400}\)…
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