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Question: For a first order reaction \( P\left( g \right) \to Q\left( g \right) + R\left( g \right) \) . After...

For a first order reaction P(g)Q(g)+R(g)P\left( g \right) \to Q\left( g \right) + R\left( g \right) . After 1010 minutes, the volume of R gas is 10L10L and after complete reaction the volume becomes 50L50L . Hence the rate constant will be (in min1{\min ^{ - 1}} )

Explanation

Solution

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 min1{\min ^{ - 1}} or sec1{\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(g)Q(g)+R(g)P\left( g \right) \to Q\left( g \right) + R\left( g \right)
It is a first order reaction.
At t=10mint = 10\min , Volume of R gas =10L= 10L
After completion of reaction, Volume of R gas = 50L50L
And we have to find the rate constant (K)\left( K \right)
P(g)Q(g)+R(g)P\left( g \right) \to Q\left( g \right) + R\left( g \right)
T= 00 aa 00 00
T=10min axa - x xx xx
00 aa aa
aa = volume of gas before the reaction has started
xx = volume of R and Q gas after 10 minutes
axa - x = volume of P after 10 minutes
It is given in the question that after 10 minutes the volume of R gas is 10L10L
x=10x = 10
It is also given that the volume of R gas after completion of reaction is 50L50L
a=50La = 50L
Now applying the values in the formula;
K=1tln(aax)K = \dfrac{1}{t}\ln \left( {\dfrac{a}{{a - x}}} \right)
K=110ln(505010)K = \dfrac{1}{{10}}\ln \left( {\dfrac{{50}}{{50 - 10}}} \right)
K=110ln(5040)K = \dfrac{1}{{10}}\ln \left( {\dfrac{{50}}{{40}}} \right)
K=110ln(1.25)K = \dfrac{1}{{10}}\ln \left( {1.25} \right) min1{\min ^{ - 1}}
Thus, the correct option is D)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.