JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
The standard free energy change \(\left(\Delta G ^{\circ}\right)\) for \(50 \%\) dissociation of \(N _{2} O _{4}\) into \(NO _{2}\) at \(27^{\circ} C\) and \(1\, atm\) pressure is \(- x\, J \,mol ^{-1}\). The value of \(x\) is \(......\) (Nearest Integer) [Given \(: R =8.31 \,J \,K ^{-1} \,mol ^{-1}, \log 1.33=0.1239\) \(\ln 10=2.3]\)
- A \(520\)
- B \(430\)
- C \(931\)
- D \(710\)
Answer & Solution
Correct Answer
(D) \(710\)
Step-by-step Solution
Detailed explanation
\(\quad \quad\quad \quad N _{2} O _{4} \quad \rightleftharpoons \quad\quad2 NO _{2}\) \(t =0 \quad\quad1 \,mol\) \(t = t\quad \quad(1-0.5) \,mol \quad \quad 0.5 \times 2\,mol\) \(\quad\quad\quad\quad=0.5 \,mol \quad\quad\quad 1\, mol\)…
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