The rate of First order reaction is 0.04 ~mol ~L^-1 ~s^-1 at 10 minutes and 0.03 ~mol ~L^-1 ~s^-1 at 20 minutes after initiation. Half life of the reaction is \underline\hspace2cm minutes. (Given \log 2=0.3010, \log 3=0.4771 )

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2024 · 30 Jan · Shift 1 · Q28 with Solution

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Correct answer: 24

Use the first-order rate law in terms of rate

For a first-order reaction,

$\text{rate} = k[A]$

Since concentration decays exponentially,

$[A]_t = [A]_0 e^{-kt}$

Therefore the rate also decays exponentially:

$r_t = r_0 e^{-kt}$

So at two different times,

$\frac{r_2}{r_1} = e^{-k(t_2-t_1)}$

Substitute the given data

At $t_1=10$ min, rate $r_1=0.04$ mol L $^{-1}$ s $^{-1}$ .

At $t_2=20$ min, rate $r_2=0.03$ mol L $^{-1}$ s $^{-1}$ .

Thus,

$\frac{0.03}{0.04} = e^{-k(20-10)}$

$\frac{3}{4} = e^{-10k}$

Taking common logarithm,

$\log\left(\frac{3}{4}\right) = -10k \log e$

Using the first-order form directly in base 10:

$k = \frac{2.303}{t_2-t_1}\log\left(\frac{r_1}{r_2}\right)$

So,

= \frac{2.303}{10}\log\left(\frac{4}{3}\right)$$ Now, $$\log\left(\frac{4}{3}\right)=\log 4-\log 3=2\log 2-\log 3$$ Given: $$\log 2=0.3010, \quad \log 3=0.4771$$ Hence, $$\log 4=2(0.3010)=0.6020$$ $$\log\left(\frac{4}{3}\right)=0.6020-0.4771=0.1249$$ Therefore, $$k=\frac{2.303}{10}(0.1249)\approx 0.0288\ \text{min}^{-1}$$ 3. **Calculate half-life** For a first-order reaction, $$t_{1/2}=\frac{0.693}{k}$$ So, $$t_{1/2}=\frac{0.693}{0.0288}\approx 24.1\ \text{min}$$ Thus the half-life is approximately $$\boxed{24\ \text{minutes}}$$ 4. **Comparison with stored answer** Stored correct answer = $24$. Our derived answer also is $24$.

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View written solutionFree

Correct answer: 24

Use the first-order rate law in terms of rate

For a first-order reaction,

$\text{rate} = k[A]$

Since concentration decays exponentially,

$[A]_t = [A]_0 e^{-kt}$

Therefore the rate also decays exponentially:

$r_t = r_0 e^{-kt}$

So at two different times,

$\frac{r_2}{r_1} = e^{-k(t_2-t_1)}$

Substitute the given data

At $t_1=10$ min, rate $r_1=0.04$ mol L $^{-1}$ s $^{-1}$ .

At $t_2=20$ min, rate $r_2=0.03$ mol L $^{-1}$ s $^{-1}$ .

Thus,

$\frac{0.03}{0.04} = e^{-k(20-10)}$

$\frac{3}{4} = e^{-10k}$

Taking common logarithm,

$\log\left(\frac{3}{4}\right) = -10k \log e$

Using the first-order form directly in base 10:

$k = \frac{2.303}{t_2-t_1}\log\left(\frac{r_1}{r_2}\right)$

So,

= \frac{2.303}{10}\log\left(\frac{4}{3}\right)$$ Now, $$\log\left(\frac{4}{3}\right)=\log 4-\log 3=2\log 2-\log 3$$ Given: $$\log 2=0.3010, \quad \log 3=0.4771$$ Hence, $$\log 4=2(0.3010)=0.6020$$ $$\log\left(\frac{4}{3}\right)=0.6020-0.4771=0.1249$$ Therefore, $$k=\frac{2.303}{10}(0.1249)\approx 0.0288\ \text{min}^{-1}$$ 3. **Calculate half-life** For a first-order reaction, $$t_{1/2}=\frac{0.693}{k}$$ So, $$t_{1/2}=\frac{0.693}{0.0288}\approx 24.1\ \text{min}$$ Thus the half-life is approximately $$\boxed{24\ \text{minutes}}$$ 4. **Comparison with stored answer** Stored correct answer = $24$. Our derived answer also is $24$.

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