Question

For a first order reaction (R→P), which is/are correct here?

• A. The time taken for the completion of 75% reaction is three times the t$_{1/2}$ of the reaction
• B. The degree of dissociation is equal to 1-e$^{-kt}$
• C. A plot of reciprocal concentration of the reactant versus time gives a straight line
• D. The pre-exponential factor in the Arrhenius equation has the dimension of Time$^{-1}$.

Answer 1

Answer: (B), (D)

(B) and (D)

Answer 2

1. Analysis of Option (A):

For a first-order reaction, the integrated rate law is given by: $k = \frac{1}{t} \ln \frac{[R]_0}{[R]}$

The half-life ($t_{1/2}$) is the time taken for the concentration of the reactant to reduce to half its initial value, i.e., $[R] = [R]_0/2$. $t_{1/2} = \frac{1}{k} \ln \frac{[R]_0}{[R]_0/2} = \frac{1}{k} \ln 2$

The time taken for 75% completion ($t_{75\%}$) means that 75% of the reactant has reacted, so the remaining concentration is $[R] = [R]_0 - 0.75[R]_0 = 0.25[R]_0 = [R]_0/4$. $t_{75\%} = \frac{1}{k} \ln \frac{[R]_0}{[R]_0/4} = \frac{1}{k} \ln 4$

Since $\ln 4 = \ln (2^2) = 2 \ln 2$, $t_{75\%} = \frac{2 \ln 2}{k}$

Comparing $t_{75\%}$ and $t_{1/2}$: $t_{75\%} = 2 \times \left(\frac{\ln 2}{k}\right) = 2 \times t_{1/2}$

So, the time taken for 75% completion is twice the half-life, not three times. Therefore, option (A) is incorrect.

2. Analysis of Option (B):

The degree of dissociation ($\alpha$) is the fraction of the reactant that has reacted. $\alpha = \frac{\text{amount reacted}}{\text{initial amount}} = \frac{[R]_0 - [R]}{[R]_0} = 1 - \frac{[R]}{[R]_0}$

From the integrated rate law for a first-order reaction, we have: $kt = \ln \frac{[R]_0}{[R]}$ Taking the exponential of both sides: $e^{kt} = \frac{[R]_0}{[R]}$ Rearranging, we get: $\frac{[R]}{[R]_0} = e^{-kt}$

Substitute this expression into the equation for $\alpha$: $\alpha = 1 - e^{-kt}$ Therefore, option (B) is correct.

3. Analysis of Option (C):

For a first-order reaction, the integrated rate law is $\ln[R] = -kt + \ln[R]_0$. A plot of $\ln[R]$ versus time ($t$) yields a straight line with a slope of $-k$ and an intercept of $\ln[R]_0$.

A plot of reciprocal concentration of the reactant ($1/[R]$) versus time ($t$) gives a straight line for a second-order reaction, where the integrated rate law is $\frac{1}{[R]} = kt + \frac{1}{[R]_0}$. Therefore, option (C) is incorrect for a first-order reaction.

4. Analysis of Option (D):

The Arrhenius equation relates the rate constant ($k$) to temperature: $k = A e^{-E_a/RT}$

Here, $A$ is the pre-exponential factor (or frequency factor), $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature.

The exponent term, $-E_a/RT$, is dimensionless. This means that the dimensions of the rate constant ($k$) must be the same as the dimensions of the pre-exponential factor ($A$).

For a first-order reaction, the rate law is: Rate $= k[R]$

The units of Rate are typically Concentration/Time (e.g., mol L$^{-1}$ s$^{-1}$). The units of Concentration ($[R]$) are Concentration (e.g., mol L$^{-1}$).

To find the units of $k$: Units of $k = \frac{\text{Units of Rate}}{\text{Units of }[R]} = \frac{\text{Concentration/Time}}{\text{Concentration}} = \text{Time}^{-1}$ (e.g., s$^{-1}$, min$^{-1}$).

Since the dimension of $k$ for a first-order reaction is Time$^{-1}$, the dimension of the pre-exponential factor $A$ must also be Time$^{-1}$. Therefore, option (D) is correct.

Conclusion:

Both options (B) and (D) are correct.