JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
For the following reaction at \(50°\)C and at \(2\) atm pressure,
\(2N_2O_5(g) \rightleftharpoons 2N_2O_4(g)+O_2(g)\)
\(N_2O_5\) is \(50\%\) dissociated.
The magnitude of standard free energy change at this temperature is \(x\).
\(x=\) ______ J mol\(^{-1}\) [Nearest integer].
Given: \(R=8.314\) J mol\(^{-1}\) K\(^{-1}\), \(\log 2=0.30\), \(\log 3=0.48\), \(\ln 10=2.303\), \(°C+273=K\)
- A 2472
- B 2473
- C 2474
- D 2475
Answer & Solution
Correct Answer
(C) 2474
Step-by-step Solution
Detailed explanation
Let the initial moles of \(N_2O_5\) be \(2\). The given reaction is: \(2N_2O_5(g) \rightleftharpoons 2N_2O_4(g) + O_2(g)\) Since \(N_2O_5\) is \(50\%\) dissociated (degree of dissociation \(\alpha = 0.5\)), the moles at equilibrium are: Moles of \(N_2O_5 = 2(1 - 0.5) = 1\) Moles…
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