For the following reaction at $50^{\circ} \mathrm{C}$ and at 2 atm pressure,
$$
2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g}) \mathrm{f} \quad 2 \mathrm{~N}_2 \mathrm{O}_4(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})
$$
$\mathrm{N}_2 \mathrm{O}_5$ is $50 \%$ dissociated.
The magnitude of standard free energy change at this temperature is $x$.
$x= \mathrm{Jmol}^{-1}$ [Nearest integer].
Given: $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 2=0.30, \log 3=0.48, \ln 10=2.303,{ }^{\circ} \mathrm{C}+273=\mathrm{K}$
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇