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Given the reaction between 2 gases represented by $A_2$ and $B_2$ to give the compound AB(g): $A_2(g) + B_2(g) \rightleftharpoons 2AB(g)$. At equilibrium, the concentration of $A_2 = 3.0\times10^{-3}$ M, of $B_2 = 4.2\times10^{-3}$ M, of AB $= 2.8\times10^{-3}$ M. If the reaction takes place in a sealed vessel at $527^\circ C$, then the value of $K_c$ will be:
A
4.5
B
2.0
C
1.9
D
0.62
Detailed Solution
$A_2 + B_2 \rightleftharpoons 2AB$
$K_c = \frac{[AB]^2}{[A_2][B_2]}$
$K_c = \frac{(2.8\times10^{-3})^2}{(3.0\times10^{-3})(4.2\times10^{-3})} = \frac{7.84}{12.6}$
$K_c = 0.62$
$K_c = \frac{[AB]^2}{[A_2][B_2]}$
$K_c = \frac{(2.8\times10^{-3})^2}{(3.0\times10^{-3})(4.2\times10^{-3})} = \frac{7.84}{12.6}$
$K_c = 0.62$