Question

For the reaction $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$, the standard free energy is $\Delta G^{\circ }>0$. The equilibrium constant $\left(K\right)$ would be

• A. $K=1$
• B. $K>1$
• C. $K<1$
• D. $K=0$

Answer 1

The correct option is C $K<1$

Here the given reaction is reversible

$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$

For a reversible reaction, at equilibrium we can define equilibrium constant $K_p$ or $K_c$.

We know that for a given chemical equilibrium,

$\Delta G=\Delta G^{\circ }+RTln{K}_c$ ---------$\left(1\right)$

Where,

$\Delta G=$ change in gibbs energy
$\Delta G^{\circ }=$ change in standard Gibb’s energy
$R=$ Gas constant
$T=$ Temperature in $K$

$K_c=$ equilibrium constant for a given equilibrium reaction.

At equilibrium, $\Delta G=0$

$\therefore$ Equation $\left(1\right)$ becomes

$0=\Delta G^{\circ }+RTln{K}_c$

i.e., $\Delta G^{\circ }=-RTln{K}_c$

$\therefore ln{K}_c=\frac{-\Delta G^{\circ }}{RT}$

As value of $\Delta G^{\circ }>0,ln{K}_c<0,K_c<1$

$\therefore$ Correct option is $\left(D\right)$.