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JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry

Dissociation of a gas \(A _2\) takes place according to the following chemical reactions. At equilibrium, the total pressure is 1 bar at 300 K .
\(A_2(g) \rightleftharpoons 2 A(g)\)
The standard Gibbs energy of formation of the involved substances has been provided below:
Substance\(\Delta G _{ t }^{\circ} / kJ mol ^{-1}\)
\(A _2\)-100.00
\(A \)-50.832
The degree of dissociation of \(A _2(g)\) is given by \(\left( x \times 10^{-2}\right)^{1 / 2}\) where \(x =\) _______.
(Nearest integer).
[Given: R =8 J \(mol ^{-1} K^{-1}\), log 2 = 0.3010, log 3 = 0.48]

  1. A 30
  2. B 33
  3. C 35
  4. D 38
Verified Solution

Answer & Solution

Correct Answer

(B) 33

Step-by-step Solution

Detailed explanation

\(-1.664 \times 10^3=-8.3 \times 300 \ln K_{ p }\) \(\ln K _{ p }=0.693\) \(K_{ p }=2\) \(2=\frac{4 \alpha^2 P _0}{1-\alpha^2}\) \(\alpha=\frac{1}{\sqrt{3}}\) \(\alpha=\left(\frac{100}{3} \times 10^{-2}\right)^{1 / 2}\) \(=\left(33.33 \times 10^{-2}\right)^{1 / 2}\)
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