Question

For a first order gas phase reaction:
\[{A_{(g)}} \to 2{B_{(g)}} + {C_{(g)}}\]
\[{P_0}\] be the initial pressure of A and P, the total pressure at time ‘t’. Integrated rate equation is:
A. \[\dfrac{{2.303}}{t}\log (\dfrac{{{P_0}}}{{{P_0} - {P_t}}})\]
B. \[\dfrac{{2.303}}{t}\log (\dfrac{{2{P_0}}}{{3{P_0} - {P_t}}})\]
C. \[\dfrac{{2.303}}{t}\log (\dfrac{{{P_0}}}{{2{P_0} - {P_t}}})\]
D. \[\dfrac{{2.303}}{t}\log (\dfrac{{2{P_0}}}{{2{P_0} - {P_t}}})\]

Answer 1

Hint: For this question we need to know the initial and final pressures and then we will find the total pressure. This will help in correct substitution of pressure values.

Complete step by step answer:
According to the reaction given,
\[{A_{(g)}} \to 2{B_{(g)}} + {C_{(g)}}\]
The initial pressures for reactants A, B and C would be \[{P_0}\] , \[0\] and \[0\] . The final pressure would be \[{P_0} - P\] , \[2P\] and \[P\] .
So, the total pressure at time t would be
\[ {P_t}= {P_0} - P + 2P + P \]
\[ \Rightarrow {P_t} = {P_0} + 2P\]
\[ \Rightarrow {P_t} - {P_0} = 2P\]
\[ \Rightarrow P = \dfrac{{{P_t} - {P_0}}}{2}\]
\[k = \dfrac{{2.303}}{t}\log [\dfrac{{{P_0}}}{{{P_0} - P}}] = \dfrac{{2.303}}{t}\log [\dfrac{{{P_0}}}{{{P_0} - (\dfrac{{{P_t} - {P_0}}}{2})}}]\]
\[ \Rightarrow \dfrac{{2.303}}{t}\log (\dfrac{{2{P_0}}}{{2{P_0} - {P_t} + {P_0}}}) = \dfrac{{2.303}}{t}\log (\dfrac{{2{P_0}}}{{3{P_0} - {P_t}}})\]

Hence, the correct option is B.

Additional Information:
Rate equation or rate law for a chemical reaction is an equation that links the initial or forward reaction rate with the concentrations or pressures of the reactants and constant parameters. There are four types of order of reaction and they are zero order, first order, second order and Pseudo first order.
First Order reaction depends on the concentration of only one reactant and it is also called unimolecular reaction. Other reactants can be present but each will be of zero order. The half-life is independent of the starting concentration. First order reactions often have the general form \[A \to products\] . The differential rate for a first order reaction is as follows:
\[rate = - \dfrac{{\Delta [A]}}{{\Delta t}} = k[A]\]
As the reaction rate is directly proportional to the concentration of one of the reactants, if the concentration of A is doubled the reaction rate doubles and if it is increased by 10 times then the reaction rate also increases by 10 times.

Note:
The equation for every order is different and therefore the use of appropriate equations is necessary as per the order of the reaction.The general equation can be obtained by solving the equation
\[rate = - \dfrac{{\Delta [A]}}{{\Delta t}} = k[A]^{\rm{n}}\]
where n is the order of the reaction.