
In a homogeneous reaction $A\to B+C+D$ the initial pressure was ${{P}_{0}}$ and after time t it was P. Expression for rate constant k in terms of ${{P}_{0}}$, P and t will be:
(A) $k=\dfrac{2.303}{t}\log \dfrac{2{{P}_{0}}}{3{{P}_{0}}-P}$
(B) $k=\dfrac{2.303}{t}\log \dfrac{2{{P}_{0}}}{{{P}_{0}}-P}$
(C) $k=\dfrac{2.303}{t}\log \dfrac{3{{P}_{0}}-P}{2{{P}_{0}}}$
(D) $k=\dfrac{2.303}{t}\log \dfrac{2{{P}_{0}}}{3{{P}_{0}}-2P}$
Answer
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Hint: In the given reaction we can see that there is only one reactant. Hence, we can assume that the reaction is a first-order reaction. In a first-order reaction, the rate of the reaction is directly dependent upon the amount of the reactant present.
Complete answer:
When the reaction takes place in the gaseous phase, we can use the partial pressure of the reactant instead of the amount of the reactant to determine the rate of the reaction.
\[k=\dfrac{2.303}{t}\log (\dfrac{{{P}^{0}}}{{{P}_{R}}})\]
Where ${{P}^{0}}$ is the initial partial pressure of the reactant and ${{P}_{R}}$ is the partial pressure of the gas at time t.
Now in the reaction
$A\to B+C+D$
The partial pressures of gases at initially and after time t has passed is:
So, the rate constant can be given as
\[k=\dfrac{2.303}{t}\log (\dfrac{{{P}^{0}}}{{{P}^{0}}-{{P}_{1}}})\]
So, the total partial pressure P at time t will be
\[\begin{align}
& P={{P}^{0}}-{{P}_{1}}+{{P}_{1}}+{{P}_{1}}+{{P}_{1}} \\
& P=2{{P}_{1}}+{{P}^{0}} \\
\end{align}\]
Hence, we can say that the value of
\[{{P}_{1}}=\dfrac{P-{{P}^{0}}}{2}\]
Substituting this value in the rate constant we get
\[\begin{align}
& k=\dfrac{2.303}{t}\log (\dfrac{{{P}^{0}}}{{{P}^{0}}-\dfrac{P-{{P}^{0}}}{2}}) \\
& k=\dfrac{2.303}{t}\log (\dfrac{2{{P}^{0}}}{3{{P}^{0}}-P}) \\
\end{align}\]
So, the correct answer is option (A).
Note:
It should be noted that this method of determination of rate constant is only applicable to first-order reactions. In all other reactions except the first-order reaction, the rate of the reaction is dependent on multiple reactants.
Also, while determining the rate constant for a reaction, the number of molecules of the gases involved in the reaction must be considered.
Complete answer:
When the reaction takes place in the gaseous phase, we can use the partial pressure of the reactant instead of the amount of the reactant to determine the rate of the reaction.
\[k=\dfrac{2.303}{t}\log (\dfrac{{{P}^{0}}}{{{P}_{R}}})\]
Where ${{P}^{0}}$ is the initial partial pressure of the reactant and ${{P}_{R}}$ is the partial pressure of the gas at time t.
Now in the reaction
$A\to B+C+D$
The partial pressures of gases at initially and after time t has passed is:
| A | B | C | D | |
| t = 0 | ${{P}^{0}}$ | 0 | 0 | 0 |
| t = t | ${{P}^{0}}-{{P}_{1}}$ | ${{P}_{1}}$ | ${{P}_{1}}$ | ${{P}_{1}}$ |
So, the rate constant can be given as
\[k=\dfrac{2.303}{t}\log (\dfrac{{{P}^{0}}}{{{P}^{0}}-{{P}_{1}}})\]
So, the total partial pressure P at time t will be
\[\begin{align}
& P={{P}^{0}}-{{P}_{1}}+{{P}_{1}}+{{P}_{1}}+{{P}_{1}} \\
& P=2{{P}_{1}}+{{P}^{0}} \\
\end{align}\]
Hence, we can say that the value of
\[{{P}_{1}}=\dfrac{P-{{P}^{0}}}{2}\]
Substituting this value in the rate constant we get
\[\begin{align}
& k=\dfrac{2.303}{t}\log (\dfrac{{{P}^{0}}}{{{P}^{0}}-\dfrac{P-{{P}^{0}}}{2}}) \\
& k=\dfrac{2.303}{t}\log (\dfrac{2{{P}^{0}}}{3{{P}^{0}}-P}) \\
\end{align}\]
So, the correct answer is option (A).
Note:
It should be noted that this method of determination of rate constant is only applicable to first-order reactions. In all other reactions except the first-order reaction, the rate of the reaction is dependent on multiple reactants.
Also, while determining the rate constant for a reaction, the number of molecules of the gases involved in the reaction must be considered.
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