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Chemistry Question on Equilibrium

Given the reaction between 2 gases represented by A2A_2 and B2B_2 to give the compound AB(g){AB_{(g)}}. A2(g)+B2(g)<=>2 AB(g){ A_{2(g)} + B_{2(g)} <=> 2 \, AB_{(g)}} At equilibrium, the concentration of A2=3.0×10−3 M{A_2 = 3.0 \times 10^{-3} \, M} of B2=4.2×10−3 M{B_2 = 4.2 \times 10^{-3}\, M } of AB=2.8×10−3 M{AB = 2.8 \times 10^{-3} \, M} If the reaction takes place in a sealed vessel at 527∘C527^{\circ} C, then the value of KCK_C will be :

A

1.9

B

0.62

C

4.5

D

2

Answer

0.62

Explanation

Solution

A2+B2<=>2AB{A_2 + B_2 <=> 2AB}
Kc=[AB]2[A2][B2]=2.8×2.8×10−63×10−3×4.2×10−3=0.62{K_{c } = \frac{[AB]^2}{[A_2][B_2]} = \frac{2.8 \times 2.8 \times 10^{-6}}{3 \times 10^{-3} \times 4.2 \times 10^{-3}} = 0.62}