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Chemistry Question on Equilibrium

The equilibrium constant for the reaction N2(g)+O2(g)⇌2NO(g)N _{2( g )}+ O _{2( g )} \rightleftharpoons 2 NO _{( g )} is 4×10−44 \times 10^{-4} at 2000 K2000\,K. In presence of a catalyst the equilibrium is attained ten times faster. Therefore the equilibrium constant in presence of catalyst of 2000 K2000\, K is

A

40×10−440 \times 10^{-4}

B

4×10−24 \times 10^{-2}

C

4×10−34 \times 10^{-3}

D

4×10−44 \times 10^{-4}

Answer

4×10−44 \times 10^{-4}

Explanation

Solution

∵K=[ conc. of product ][ conc. of reactant ]\because K =\frac{[\text { conc. of product }]}{[\text { conc. of reactant }]}
=4×10−4=4 \times 10^{-4}

Qn adding catalyst. the equilibrium attain 10 time factor.

As equilibrium constant depends on temperature. It will not change, however the reaction attains equilibrium 10 times faster hence

K_{\text {(new) }}+K_{\text {(previous\\} }}=4 \times 10^{-4}