Question

For first order kinetics, Co is the initial concentration. If 25% of reactant decomposes in t25% then, concentration of reactant left after n t25%

• A. $({\frac{3}{2}})^n$ Co
• B. $\frac{3^n}{2^{2n}}$Co
• C. $({\frac{3}{2}})^{2n}$ Co
• D. $({\frac{3}{4}})^{n-1}$ Co

Answer 1

Answer: (B)

$\frac{3^n}{2^{2n}}$Co

Answer 2

The problem asks for the concentration of a reactant left after $n \cdot t_{25\%}$ for a first-order reaction, where $t_{25\%}$ is the time taken for 25% of the reactant to decompose.

1. Integrated Rate Law for First-Order Reactions:

For a first-order reaction, the integrated rate law is given by:

$kt = \ln\left(\frac{C_0}{C_t}\right)$

where:

• $k$ is the rate constant
• $t$ is the time
• $C_0$ is the initial concentration
• $C_t$ is the concentration at time $t$

Alternatively, it can be written as:

$C_t = C_0 e^{-kt}$

2. Determine the relationship involving $t_{25\%}$:

Given that 25% of the reactant decomposes in $t_{25\%}$, it means that 75% of the reactant remains.

So, at $t = t_{25\%}$, the concentration remaining $C_{t_{25\%}}$ is $0.75 C_0$.

$C_{t_{25\%}} = \frac{3}{4} C_0$

Substitute this into the integrated rate law:

$k \cdot t_{25\%} = \ln\left(\frac{C_0}{\frac{3}{4} C_0}\right)$

$k \cdot t_{25\%} = \ln\left(\frac{4}{3}\right)$

3. Calculate the concentration left after $n \cdot t_{25\%}$:

Let the total time be $T = n \cdot t_{25\%}$. We want to find the concentration $C_T$ at this time.

Using the integrated rate law:

$k \cdot T = \ln\left(\frac{C_0}{C_T}\right)$

Substitute $T = n \cdot t_{25\%}$:

$k \cdot (n \cdot t_{25\%}) = \ln\left(\frac{C_0}{C_T}\right)$

$n \cdot (k \cdot t_{25\%}) = \ln\left(\frac{C_0}{C_T}\right)$

Now, substitute the value of $(k \cdot t_{25\%})$ from step 2:

$n \cdot \ln\left(\frac{4}{3}\right) = \ln\left(\frac{C_0}{C_T}\right)$

Using the logarithm property $a \ln b = \ln b^a$:

$\ln\left(\left(\frac{4}{3}\right)^n\right) = \ln\left(\frac{C_0}{C_T}\right)$

Since the natural logarithms are equal, their arguments must be equal:

$\left(\frac{4}{3}\right)^n = \frac{C_0}{C_T}$

To find $C_T$, rearrange the equation:

$C_T = C_0 \cdot \frac{1}{\left(\frac{4}{3}\right)^n}$

$C_T = C_0 \cdot \left(\frac{3}{4}\right)^n$

The derived expression matches option B.