Q.57. Consider the first order reaction R \rightarrow P. The fraction of molecules decomposed in the

Q: What is the correct answer for this exam Chemistry Q57?

The correct answer is D. For a first-order reaction, the integrated rate law is $k t = \ln(\frac{[R]_0}{[R]_t})$, where $[R]_0$ is the initial concentration and $[R]_t$ is the concentration at time $t$.Let $[R]_0 = a$ and $[R]_t = a-x$, where $x$ is the amount decomposed.So, $kt = \ln(\frac{a}{a-x})$.Exponentiating both sides: $e^{kt} = \frac{a}{a-x}$.Rearranging: $a-x = a e^{-kt}$.So, $x = a - a e^{-kt} = a(1 - e^{-kt})$

Q: Q.57. Consider the first order reaction R \rightarrow P. The fraction of molecules decomposed in the given first order reaction can be expressed as

The correct answer is D. For a first-order reaction, the integrated rate law is $k t = \ln(\frac{[R]_0}{[R]_t})$, where $[R]_0$ is the initial concentration and $[R]_t$ is the concentration at time $t$.Let $[R]_0 = a$ and $[R]_t = a-x$, where $x$ is the amount decomposed.So, $kt = \ln(\frac{a}{a-x})$.Exponentiating both sides: $e^{kt} = \frac{a}{a-x}$.Rearranging: $a-x = a e^{-kt}$.So, $x = a - a e^{-kt} = a(1 - e^{-kt})$

Q: What are the options for this this exam Chemistry question?

A. $1-e^{kt}$ · B. $1+e^{kt}$ · C. $1+e^{-kt}$ · D. $1-e^{-kt}$

Question

Q.57. Consider the first order reaction R \rightarrow P. The fraction of molecules decomposed in the given first order reaction can be expressed as

FirstInTest · Accepted Answer

For a first-order reaction, the integrated rate law is $k t = \ln(\frac{[R]_0}{[R]_t})$, where $[R]_0$ is the initial concentration and $[R]_t$ is the concentration at time $t$.Let $[R]_0 = a$ and $[R]_t = a-x$, where $x$ is the amount decomposed.So, $kt = \ln(\frac{a}{a-x})$.Exponentiating both sides: $e^{kt} = \frac{a}{a-x}$.Rearranging: $a-x = a e^{-kt}$.So, $x = a - a e^{-kt} = a(1 - e^{-kt})$.The fraction of molecules decomposed is $\frac{x}{a} = \frac{a(1 - e^{-kt})}{a} = 1 - e^{-kt}$.

Answer

$1-e^{kt}$

Answer

$1+e^{kt}$

Answer

$1+e^{-kt}$

Answer

$1-e^{-kt}$

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