Question

For a first order reaction $ P\left( g \right) \to Q\left( g \right) + R\left( g \right) $ . After $ 10 $ minutes, the volume of R gas is $ 10L $ and after complete reaction the volume becomes $ 50L $ . Hence the rate constant will be (in $ {\min ^{ - 1}} $ )

Answer 1

Hint :Rate constant is the proportionality constant, that relates rate of the reaction with the molar concentration of the reactants. For a first order reaction, the unit of rate constant is $ {\min ^{ - 1}} $ or $ {\sec ^{ - 1}} $ The unit of rate constant is different for zero and second order reaction.

Complete Step By Step Answer:
In the question, the reaction given is;
 $ P\left( g \right) \to Q\left( g \right) + R\left( g \right) $
It is a first order reaction.
At $ t = 10\min $ , Volume of R gas $ = 10L $
After completion of reaction, Volume of R gas = $ 50L $
And we have to find the rate constant $ \left( K \right) $
              $ P\left( g \right) \to Q\left( g \right) + R\left( g \right) $
T= $ 0 $ $ a $ $ 0 $ $ 0 $
T=10min $ a - x $ $ x $ $ x $
                       $ 0 $ $ a $ $ a $
 $ a $ = volume of gas before the reaction has started
 $ x $ = volume of R and Q gas after 10 minutes
 $ a - x $ = volume of P after 10 minutes
It is given in the question that after 10 minutes the volume of R gas is $ 10L $
 $ x = 10 $
It is also given that the volume of R gas after completion of reaction is $ 50L $
 $ a = 50L $
Now applying the values in the formula;
 $ K = \dfrac{1}{t}\ln \left( {\dfrac{a}{{a - x}}} \right) $
 $ K = \dfrac{1}{{10}}\ln \left( {\dfrac{{50}}{{50 - 10}}} \right) $
 $ K = \dfrac{1}{{10}}\ln \left( {\dfrac{{50}}{{40}}} \right) $
 $ K = \dfrac{1}{{10}}\ln \left( {1.25} \right) $ $ {\min ^{ - 1}} $
Thus, the correct option is $ D) $ .

Additional Information:
The interesting thing to note in first order reactions is that the half-life of first order reactions is independent of the concentration of the reactant. While, the zero and the second order reactions are dependent on initial concentration of the reactant. Order of a reaction is experimentally determined and is not equal to the stoichiometry except in elementary reactions.

Note :
In such questions, firstly one should find the order of the reaction and then solve the problem. We can recognise the order from the units of the rate constant. Also, in many cases the order of the reaction is given in the question itself.