Question

In this case we have (Consider it as a first-order reaction)
$A \to B + C$
\[
  \begin{array}{*{20}{c}}
  {Time}&{{\text{ }}t}&\infty
\end{array} \\
  \begin{array}{*{20}{c}}
  {{\text{Total pressure}}}&{{P_2}}&{{P_3}}
\end{array} \\
 \]
Then K is.
A) $K = \dfrac{1}{t}\ln \dfrac{{{p_3}}}{{2\left( {{p_3} - {p_2}} \right)}}$
B) $K = \dfrac{1}{t}\ln \dfrac{{{p_2}}}{{2\left( {{p_3} - {p_2}} \right)}}$
C) $K = \dfrac{1}{t}\ln \dfrac{{{p_3}}}{{2\left( {{p_1} - {p_2}} \right)}}$
D) $K = \dfrac{1}{t}\ln \dfrac{{{p_1}}}{{2\left( {{p_3} - {p_2}} \right)}}$

Answer 1

Hint: We know that the rate of order is determined by the differential rate law or the integrated rate law, rate of reaction is speed of chemical reaction proceeds measure of the change in concentration product or the change in concentration of the reactants.
If the kinetics of the reaction depends on the concentration of only one reactant then the order of the reaction is first order present, but each reactant will be zeroth-order.

Complete step by step answer:
We need to remember that the integrated rate equations are widely used to calculate the rate constant of a reaction by experimentally. Differential rate law for first order reaction can be modified as,
$\dfrac{{ - d\left[ A \right]}}{{dt}} = k\left[ A \right]$
$ \Rightarrow \dfrac{{d\left[ A \right]}}{{\left[ A \right]}} = - kdt$
On integration of both sides we get,
$\int\limits_{{{\left[ A \right]}_0}}^{\left[ A \right]} {\dfrac{{d\left[ A \right]}}{{\left[ A \right]}}} = - \int\limits_{{t_0}}^t {kdt} $
We can write the above equation as,
$\int\limits_{{{\left[ A \right]}_0}}^{\left[ A \right]} {\dfrac{1}{{\left[ A \right]}}} d\left[ A \right] = - \int\limits_{{t_0}}^t {kdt} $
As we know that, $\int {\dfrac{1}{x}} = \ln \left( x \right)$ , so the equation becomes,
$\ln \left[ A \right] - \ln {\left[ A \right]_0} = - kt$
$ \Rightarrow \ln \left[ A \right] = - kt + \ln {\left[ A \right]_0}$
Or
$\ln \left[ A \right] = \ln {\left[ A \right]_0} - kt$
We can transform the above equation by exponent ‘e’, we get,
${e^{\ln \left[ A \right]}} = {e^{\ln {{\left[ A \right]}_0} - kt}}$
As we know that, ${e^{\ln \left( X \right)}} = X$
The integrated rate law for a first-order reaction is frequently written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:
\[\left[ A \right] = {\left[ A \right]_0}{e^{ - kt}}\]
Initial concentration of reactant A at \[t = 0\] is denoted as $\left[ {{A_0}} \right]$ ,
K is that the rate constant, and e is that the base of the natural logarithms, which has the worth \[2.718\] to $3$ decimal places. By taking the logarithm of each side of the equation and rearranging, we get an alternate logarithmic expression of the connection between the concentration of A and t.
\[ln\left[ A \right] = ln{\left[ A \right]_0}{e^{ - k}}^t\]
If time t is infinite A has reacted entirely and one mole of A gives one mole of B and one mole of C.
$a = \dfrac{{{p_3}}}{2}$
The total pressure is $a - x + x + x = a + x = {p_2}$
Thus, $x = {p_2} - a = 2a - {p_2} = {p_3} - {p_2}$
The rate law for a first-order reaction is $k = \dfrac{1}{t}\ln \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}} = \dfrac{1}{t}\ln \dfrac{{{p_3}}}{2}/\left( {{p_3} - {p_2}} \right)$
The rate law for a first-order reaction is $k = \dfrac{1}{t}\ln \dfrac{{{p_3}}}{{2\left( {{p_3} - {p_2}} \right)}}$.

So, the correct answer is Option A.

Note: For zeroth-order kinetics reaction is independent of concentration of reaction, the changing of concentration does not impact the rate of reaction.
For second-order kinetics the sum of the order is equal to two in the rate law of chemical reaction.