Question

$A(g)+3B(g)4C(g)$ Initially concentration of $A$ is equal to that of $B$ . The equilibrium concentrations of $A$ and $C$ are equal. ${{K}_{c}}$ is:

• A. 0.08
• B. 0.8
• C. 8
• D. 80

Answer 1

Answer: (C)

8

Answer 2

$\underset{\begin{smallmatrix} \,\,\,\,\,1 \\\ (1-x) \end{smallmatrix}}{\mathop{A(g)}}\,+\underset{\begin{smallmatrix} \,\,\,\,\,1 \\\ (1-3x) \end{smallmatrix}}{\mathop{3B(g)}}\,\underset{\begin{smallmatrix} 0 \\\ 4x \end{smallmatrix}}{\mathop{4C(g)}}\,\underset{\begin{smallmatrix} (initial\text{ }concentration) \\\ (final\text{ }concentration) \end{smallmatrix}}{\mathop{{}}}\,$ (at equilibrium)
According to question, $1-x=4x$ $\therefore$ $x=\frac{1}{5}$
For above reaction ${{K}_{c}}=\frac{{{[C]}^{4}}}{[A]{{[B]}^{3}}}=\frac{{{(4x)}^{4}}}{(1-x){{(1-3x)}^{3}}}$ ${{K}_{c}}=\frac{{{\left( 4\times \frac{1}{5} \right)}^{4}}}{\left( 1-\frac{1}{5} \right){{\left( 1-3\times \frac{1}{5} \right)}^{3}}}=8.0$