Question

Calculate the equilibrium concentration ratio of $CtoA$, if $3$ mol each of $A$ and $B$ are allowed to come to equilibrium at $298K$.
$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right)$
$\Delta G^{\circ }=-1380cal/mol$

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

Answer 1

The correct option is C $3.16$

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,
$\Delta G=0,Q=K_c\therefore \Delta G^{\circ }=-2.303RT\log K_c-1380=-2.303\times 2\times 298\times \log K_c\log K_c=\frac{-1380}{-2.303\times 2\times 298}=\frac{1380}{1373}\Rightarrow \log K_c\approx 1\therefore \log K_c=1K_c={10}^1=10$

Given reaction :
$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right)Initial\left(mol\right)3300Equilibrium:3-\alpha 3-\alpha \alpha \alpha K_c=\frac{\left[C{]}_{eq}\right[D{]}_{eq}}{\left[A{]}_{eq}\right[B{]}_{eq}}$

Taking the volume V
Equilibrium ratio of C to A is :
$\frac{[C{]}_{eq}}{[A{]}_{eq}}=\frac{\left[\frac{\alpha }{V}\right]}{\left[\frac{(3-\alpha )}{V}\right]}=\frac{\alpha }{3-\alpha }$
$K_c=\frac{{\left[\frac{\alpha }{V}\right]}^2}{{\left[\frac{(3-\alpha )}{V}\right]}^2}=\frac{{\alpha }^2}{(3-\alpha {)}^2}10=\frac{{\alpha }^2}{(3-\alpha {)}^2}\sqrt{10}=\frac{\alpha }{3-\alpha }=\frac{[C{]}_{eq}}{[A{]}_{eq}}\frac{[C{]}_{eq}}{[A{]}_{eq}}=\sqrt{10}=3.16$