Question

For the reaction $A+2B\rightleftharpoons C+D,$ the equilibrium constant is $1.0\times {10}^8$. Calculate the equilibrium concentration of A if 1.0 mole of A and 3.0 mole of B are placed in 1L flask and allowed to attain the equilibrium ?

• A. $1.0\times {10}^{-2}mol{L}^{-1}$
• B. $2.1\times {10}^{-4}mol{L}^{-1}$
• C. $5\times {10}^{-5}mol{L}^{-1}$
• D. $1.0\times {10}^{-8}mol{L}^{-1}$

Answer 1

The correct option is D

$1.0\times {10}^{-8}mol{L}^{-1}$

$A+2B\rightleftharpoons C+D;x=$ mole of A remaining at equilibrium
Moles of A reacted = 1 - x;
Moles of B reacted = 2(1 - x)
Moles of B remaining $=3-2(1-x)=1+2x\approx 1$
$(\because$ K is very large, x < < 1 )
Moles of C = Moles of $D=1-x\approx 1$
Hence, $K=1.0\times {10}^8=\frac{\left[C\right]\left[D\right]}{\left[A\right][B{]}^2}=\frac{1\times 1}{x\times [1+2x{]}^2}$
$=\frac{1}{x}\Rightarrow x=1.0\times {10}^{-8}$