Question

$A\left(g\right)+B\left(g\right)\leftrightharpoons C\left(g\right)+D\left(g\right)△G^0=2763.6cal.$
when two mol of each A and B are present initially and the mixture attains an equilibrium at $300K$, the equilibrium concentration ratio of C to A is .

Answer 1

Given that $△G^0=-2.303\times R\times T\times logK=2763.6cal$
$-2763.6=2.303\times 2\times 300\times logK$
$logK=2$
$K=100\left(1\right)$
at equilibrium:
$K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$
Because we started with 2 moles each of A and B, the number of moles of C and D formed will be equal.
$\left[A\right]=\left[B\right]and\left[C\right]=\left[D\right]\therefore K=\frac{[C{]}^2}{[A{]}^2}$
$100=\frac{[C{]}^2}{[A{]}^2}$
$\frac{\left[C\right]}{\left[A\right]}=10$