Question

For the reaction $A\left(g\right)\rightleftharpoons B\left(g\right)at495K,\Delta r{G}^o=-9.478kJmo{l}^{-1}$. If we start reaction in a closed container at 495 K with 22 millimoles of A, the amount of B is the equilibrium mixture is .....millimoles. (Round off to the Nearest integer). [R$8.314Jmo{l}^{-1}{K}^{-1};$]~ln10=2.303]

Answer 1

$\Delta G^o=-RTln{K}_{eq}$

$\Delta G^o=-9.478kJ/mole$

$T=495KR=8.314Jmo{l}^{-1}$

$-9.478\times {10}^3=-495\times 8.314\times ln{K}_{eq}$

$ln{K}_{eq}=2.303$

$ln{K}_{eq}=ln10$

$So{K}_{eq}=10$

$NowA\left(g\right)\rightleftharpoons B\left(g\right)$

$t=0220$

$t=t22-xx$

$K_{eq}=\frac{\left[B\right]}{\left[C\right]}=\frac{x}{22-x}=10$
or x =20
So millmoles of B = 20