Question

the equilibrium constant, ( k_{c} ), for the following reaction is ( 6.66 \times 10^{-3} ) at ( 299 mathrm{~k} ).
( 2 mathrm{nobr}(mathrm{g})
ightleftharpoons 2 mathrm{no}(mathrm{g})+mathrm{br}_{2}(mathrm{~g}) )
calculate ( k_{c} ) at this temperature for the following reaction:
( mathrm{nobr}(mathrm{g})
ightleftharpoons mathrm{no}(mathrm{g})+1 / 2 mathrm{br}_{2}(mathrm{~g}) )
( k_{c}= )

Explanation:

Step1: Write \(K_{c1}\) for the first reaction

For \(2NOBr(g)
ightleftharpoons2NO(g)+Br_{2}(g)\), \(K_{c1}=\frac{[NO]^{2}[Br_{2}]}{[NOBr]^{2}} = 6.66\times 10^{-3}\)

Step2: Write \(K_{c2}\) for the second reaction

For \(NOBr(g)
ightleftharpoons NO(g)+\frac{1}{2}Br_{2}(g)\), \(K_{c2}=\frac{[NO][Br_{2}]^{\frac{1}{2}}}{[NOBr]}\)

Step3: Relate \(K_{c2}\) to \(K_{c1}\)

Notice that \(K_{c2}=\sqrt{K_{c1}}\). Substitute \(K_{c1}=6.66\times 10^{-3}\) into the formula.
\(K_{c2}=\sqrt{6.66\times 10^{-3}}\)
\(K_{c2}\approx0.0816\)

Answer:

\(0.0816\)