Chemistry - Thermodynamics - II Question with Solution | TestHub

Let assume initial moles is 1
${\mathrm{ \; \; \; \; \; \; \; A}}_2\stackrel{ }{\to } 2A$
t=0       1
t=t$1-0.2 \mathrm{ 0}.2\times 2$

$Partial pressure of A_2=\frac{0.8}{1.2}\times 1$
$Partial pressure of A=\frac{0.4}{1.2}\times 1$
$k_p=\frac{{\left(\frac{0.4}{1.2}\right)}^2\times 1}{\left(\frac{0.8}{1.2}\right)\times 1 }=\frac{1}{6}$
$∆G°= -RTlnk=\mathrm{-8.314\times 320} ln\frac{1}{6}$
$=\mathrm{8.314\times 320} \left(0.693+1.098\right)$
$\backsimeq 4763 J/mole$