Question

For a homogeneous gaseous reaction, $2A\left(g\right)+2B\left(g\right)\rightleftharpoons 4C\left(g\right)+D\left(g\right)$$K_{eq}=\frac{8}{(1.5{)}^3}$
If in a 2 litre rigid container starting with 4 moles of A & 6 moles of B equilibrium was established then identify the option(s} which is/are correct.

• A. Concentration of B at equilibrium is 1.5 M
• B. Concentration of D at equilibrium is 1 M
• C. Concentration of A at equilibrium is 1 M
• D. Concentration of C at equilibrium is 2 M

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Concentration of B at equilibrium is 1.5 M
B Concentration of D at equilibrium is 1 M
C Concentration of A at equilibrium is 1 M
D Concentration of C at equilibrium is 2 M

$\underset{4moles}{2A\left(g\right)}+\underset{6moles}{3B\left(g\right)}\rightleftharpoons 4C\left(g\right)+D\left(g\right)4-2x6-3x4xx{K}_{eq}=\frac{8}{(1.5{)}^3}{K}_{eq}=\frac{(4x{)}^{4}x}{(4-2x{)}^2(6-3x{)}^3}\times \frac{\frac{1}{V^4}\times \frac{1}{V}}{\frac{1}{V^2}\times \frac{1}{V^3}}\Rightarrow \frac{8}{(1.5{)}^3}=\frac{4^4{x}^5}{4(2-x{)}^{2}27(2-x{)}^3}\Rightarrow \frac{8\times 4\times 27}{\left(1.5{)}^3\right(4{)}^4}={\left(\frac{x}{2-x}\right)}^5\Rightarrow {\left(\frac{x}{2-x}\right)}^5=1\frac{x}{2-x}=1\Rightarrow x=1[B{]}_{eq}=\frac{6-3x}{V}=\frac{6-3}{2}=\frac{3}{2}=1.5M[D{]}_{eq}=\frac{x}{V}=\frac{1}{2}=0.5M[A{]}_{eq}=\frac{4-2x}{V}=\frac{2}{2}=1M[C{]}_{eq}=\frac{4x}{V}=\frac{4\times 1}{2}=2M$