Question

In a reaction $A+2B\leftrightharpoons 2C;2molesofA,3$ mole of $Band2moleofC$ are placed in a 2 L flask and the equilibrium concentration of $Cis0.5mole/L.$ The equilibrium constant K for the reaction is:

• A. 0.073
• B. 0.147
• C. 0.05
• D. 0.026

Answer 1

The correct option is C 0.05

$\begin{array}{cccc}& A+& 2B\leftrightharpoons & 2C\\ Initialmoles& 2& 3& 2\\ Molarconc.& 2/2=1mol/L& 3/2=1.5mol/L& 2/2=1mol/L\\ AtEqui.& & & \\ Conc& 1+0.25=1.25mol/L& 1.5+0.5=2mol/L& 0.5mol/L\end{array}$

$K=\left(0.5{)}^2/\right(1.25\times \left(2{)}^2\right)=0.05$ Note that concentration of C has decreased, i.e., to attain equilibrium, reaction has shifted backwards. Decrease in conc. of C will be $0.5mol/L$ and hence increase in concentration of B will be $0.5mol/L$ and that of A will be $0.25mol/L$