Ncert Exercises & Solutions

For the reaction $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ , the standard free energy is $∆ G^{⊝} > 0$ . The equilibrium constant ( $K_{c}$ ) would be :

1.	$K_{c}$=0	2.	$K_{c}$>1
3.	$K_{c}$=1	4.	$K_{c}$<1

Hint:

∆ G^{⊝} = - RT

\ln

K_{c}

Step 1:

The relation between

∆ G^{⊝}

and

K_{c}

is as follows:

∆ G^{⊝} = - RT

\ln

K_{c}

Step 2:

when

∆ G^{⊝} > 0

means

∆ G^{⊝}

is positive. This can be so only if in

K_{c}

is negative i.e.,

K_{c} < 1

1.	\(K_{c}\)=0	2.	\(K_{c}\)>1
3.	\(K_{c}\)=1	4.	\(K_{c}\)<1