Question

Calculate the equilibrium concentration ratio of $CtoA$, if $2$ moles each of $A$ and $B$ are allowed to come to equilibrium at $300K$.
$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right)$
$\Delta G^{\circ }=-1380cal/mol$

• A. $0.56$
• B. $4.45$
• C. $1.50$
• D. $3.16$

Answer 1

The correct option is D $3.16$

We know ,
$\Delta G=\Delta G^{\circ }+2.303RTlogQ$
where:
$\Delta G,\Delta G^{\circ }$ are Gibbs free energy and standard gibbs free energy respectively.
R is the universal gas constant
T is the temperature
Q is the reaction quotient at time t
At equilibrium,
$\Delta G=0,Q=K_c\therefore \Delta G^{\circ }=-2.303RT\log K_c-1380=-2.303\times 2\times 300\times \log K_c\log K_c=\frac{-1380}{-2.303\times 2\times 300}=\frac{2300}{2303}\Rightarrow \log K_c=0.99\approx 1\therefore \log K_c=1K_c={10}^1=10$

Given reaction :
$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right)Initial\left(mol\right)2200Equilibrium:2-\alpha 2-\alpha \alpha \alpha K_c=\frac{\left[C{]}_{eq}\right[D{]}_{eq}}{\left[A{]}_{eq}\right[B{]}_{eq}}$

Taking the volume V
Equilibrium ratio of C to A is :
$\frac{[C{]}_{eq}}{[A{]}_{eq}}=\frac{\left[\frac{\alpha }{V}\right]}{\left[\frac{(2-\alpha )}{V}\right]}=\frac{\alpha }{2-\alpha }$
$K_c=\frac{{\left[\frac{\alpha }{V}\right]}^2}{{\left[\frac{(2-\alpha )}{V}\right]}^2}=\frac{{\alpha }^2}{(2-\alpha {)}^2}10=\frac{{\alpha }^2}{(2-\alpha {)}^2}\sqrt{10}=\frac{\alpha }{2-\alpha }=\frac{[C{]}_{eq}}{[A{]}_{eq}}\frac{[C{]}_{eq}}{[A{]}_{eq}}=\sqrt{10}=3.16$