\begin {aligned}
3A_{(l)}+2B_{(l)}\rightleftharpoons C_{(l)}+2D_{(l)}
\end {aligned}A mixture of 2 mole of A and 1 mole of B was allowed to reach equilibrium in a 1L vessel according to the following equation: The equilibrium mixture was found to contain 0.4 moles of B. Calculate the value for Kc of this reaction. In such a question the first two things to write are always the ICE table and the kc equation:
\begin {aligned}
k_c=\frac{[D]^2[C]}{[B]^2[A]^3}
\end {aligned}\begin {aligned}
1-2x = 0.4
\end {aligned}\begin {aligned}
x= \mathbf{{\color{Blue} 0.3}}
\end {aligned}\begin {aligned}
k_c=\frac{[0.6]^2[0.3]}{[0.4]^2[1.1]^3}
\end {aligned}The ICE table would be:
Since at equilibrium it is found that the number of moles of B = 0.4, then: This would create the number of moles of each component at equilibrium as:
Since the volume is 1L, then the concentration is the same as the number of moles. Once the concentrations are found, then it is just a matter of placing each value in the kc equation as follows: where kc would be found to be: 0.507mol-2 dm6