Question

The equlibrium constant K, for the reaction $N_2+3H_2\rightleftharpoons 2N{H}_3$ is $1.64\times {10}_{4}atm$ at ${400}^{\circ }C$ . What will be the equilibrium constant at ${500}^{\circ }C$, if heat of reaction in this temperature range is - 105185.8 Joules.

• A. 0.144 × 10-4atm
• B. 4.144 × 10-3atm
• C. 0.144 × 10-2atm
• D. 0.144 × 10-1atm

Answer 1

The correct option is A

0.144 × 10-4atm

$K_{p_1}=1.64\times {10}^{-4}atm,K_{p_2}=?$

$T_1=400+273=673K$

$T_2=500+273=773K$

$△H=-105185.8Joules$

$R=8.314J/K/mole$

Applying equation

$Log{K}_{p_2}=log1.64\times {10}^{-4}$

$\frac{105185.8}{2.303\times 8.314}(\frac{773-673}{773\times 673})$

or $K_{p_2}=0.144\times {10}^{-4}atm$