Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2007

\[1.1\]moles of A and \[2.2\] moles of B are mixed in a container of one litre volume to obtain the equilibrium \[A+2B\rightleftharpoons 2C+D.\].     At equilibrium 0.2 moles of C are formed. The equilibrium constant for the above reaction is

A) \[0.001\]

B) \[0.002\]

C) \[0.003\]

D) \[0.004\]

Correct Answer: A

Solution :

The given reaction is

	\[A+\]	\[2B\rightleftharpoons \]	\[2C+\]	\[D\]
Initial	\[a\,mol\]	\[2b\,mol\]	\[0\]	\[0\]
	\[1.1\,mol\]	\[2.2\,mol\]	\[0\]	\[0\]
At equilibrium	\[(a-x)\]	\[(2b-2x)\]	\[2x\]	\[x\]

At equilibrium \[0.2\text{ }mole\]of C formed \[\therefore \] \[2\text{ x}=0.2\text{ }mole\] \[\therefore \] \[x=0.1\text{ }mole\] Mole at equilibrium \[\underset{1\,\,mol}{\mathop{\text{(1}\text{.1-0}\text{.1) }mol}}\,\text{ }\underset{2\,\,mol}{\mathop{\text{(}2.2-0.2)\text{ }mol}}\,\text{ }\underset{0.2\,\,mol}{\mathop{0.2\text{ }mol}}\,\text{ }\underset{0.1\,mol}{\mathop{0.1\text{ }mol}}\,\] Since, volume of container is one litre. So, active mass of \[A=\text{1 }mol/L\] active mass of \[B=2\text{ }mol/L\] active mass of \[C=0.2\text{ }mol/L\] active mass of \[D=0.1\text{ }mol/L\] On applying law of mass action. \[{{K}_{c}}=\frac{{{[C]}^{2}}[D]}{[A]{{[B]}^{2}}}=\frac{{{[0.2]}^{2}}\times 0.1}{1\times {{[2]}^{2}}}\] \[\therefore \] \[{{K}_{c}}=0.001\]

Report Error