Question

use the references to access important values if needed for this question.
the equilibrium constant, ( k_{c} ), for the following reaction is ( 3.85 \times 10^{-3} ) at ( 272 mathrm{~k} ).
2 mathrm{nobr}(g)
ightleftharpoons 2 mathrm{no}(g)+mathrm{br}_{2}(g)
calculate ( k_{c} ) at this temperature for the following reaction:
mathrm{no}(g)+1 / 2 mathrm{br}_{2}(g)
ightleftharpoons mathrm{nobr}(g)
( k_{c}= )

Explanation:

Step1: Reverse the first reaction

When we reverse the reaction \(2NOBr(g)
ightleftharpoons2NO(g)+Br_{2}(g)\), the equilibrium constant \(K_{c1}\) becomes \(\frac{1}{K_{c}}\). So, \(K_{c1}=\frac{1}{3.85\times 10^{- 3}}\)

Step2: Divide the reversed reaction by 2

If we divide the reaction \(2NO(g)+Br_{2}(g)
ightleftharpoons2NOBr(g)\) (reversed of the original) by 2 to get \(NO(g)+\frac{1}{2}Br_{2}(g)
ightleftharpoons NOBr(g)\), the new equilibrium constant \(K_{c2}\) is \(\sqrt{K_{c1}}\).

So, \(K_{c2}=\sqrt{\frac{1}{3.85\times 10^{-3}}}\)

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Answer:

\(16.1\)