- A$-2.477 \,kJ$
- ✓$-4.955\, kJ$
- C$0.0489\, kJ$
- D$2.477\, kJ$
$For\,2\,mol\,\Delta n = 2 - 0 = 2$
${\text{w}} = - \Delta {\text{nRT}}$
$ = - 2 \times 8.31 \times 298\,{\text{J}}$
$ = - 4952.76\,{\text{J}}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$Cu ^{2+}+ NH _{3} \stackrel{ K _{1}}{\rightleftharpoons}\left[ Cu \left( NH _{3}\right)\right]^{2+}$
$\left[ Cu \left( NH _{3}\right)\right]^{2+}+ NH _{3} \stackrel{ K _{2}}{\rightleftharpoons}\left[ Cu \left( NH _{3}\right)_{2}\right]^{2+}$
$\left[ Cu \left( NH _{3}\right)_{2}\right]^{2+}+ NH _{3} \stackrel{ K _{3}}{\rightleftharpoons}\left[ Cu \left( NH _{3}\right)_{3}\right]^{2+}$
$\left[ Cu \left( NH _{3}\right)_{3}\right]^{2+}+ NH _{3} \stackrel{ K _{4}}{\rightleftharpoons}\left[ Cu \left( NH _{3}\right)_{4}\right]^{2+}$
The value of stability constants $K _{1}, K _{2}, K _{3}$ and $K _{4}$ are $10^{4}, 1.58 \times 10^{3}, 5 \times 10^{2}$ and $10^{2}$ respectively. The overall equilibrium constants for dissociation of $\left[ Cu \left( NH _{3}\right)_{4}\right]^{2+}$ is $x \times 10^{-12}$ The value of $x$ is ...............
(Rounded off to the nearest integer)