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Chemical Kinetics and Nuclear Chemistry question

2021 · 22 Jul · Shift 2 · Q22
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Chemical Kinetics and Nuclear Chemistry question

2021 · 22 Jul · Shift 2 · Q22

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
N2O5(g)→2NO2(g)+12O2(g){N_2}{O_{5(g)}} \to 2N{O_{2(g)}} + {1 \over 2}{O_{2(g)}}N2​O5(g)​→2NO2(g)​+21​O2(g)​ In the above first order reaction the initial concentration of N2O5N_2O_5N2​O5​ is 2.40 ×\times× 10 −-− 2 mol L −-− 1 at 318 K. The concentration of N2O5N_2O_5N2​O5​ after 1 hour was 1.60 ×\times× 10 −-− 2 mol L −-− 1. The rate constant of the reaction at 318 K is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 3 min −-− 1. (Nearest integer) [Given : log 3 = 0.477, log 5 = 0.699]
Numerical answer
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Correct answer: 7

  1. Given data

For the first-order reaction N2O5(g)→2NO2(g)+12O2(g)N_2O_5(g) \to 2NO_2(g) + \frac{1}{2}O_2(g)N2​O5​(g)→2NO2​(g)+21​O2​(g) we have:

  • Initial concentration: [A]0=2.40×10−2 mol L−1[A]_0 = 2.40 \times 10^{-2}\ \text{mol L}^{-1}[A]0​=2.40×10−2 mol L−1
  • Concentration after 111 hour: [A]t=1.60×10−2 mol L−1[A]_t = 1.60 \times 10^{-2}\ \text{mol L}^{-1}[A]t​=1.60×10−2 mol L−1
  • Time: t=1 hour=60 mint = 1\ \text{hour} = 60\ \text{min}t=1 hour=60 min

We need the rate constant kkk in the form: ‾×10−3 min−1\underline{\hspace{1cm}} \times 10^{-3}\ \text{min}^{-1}​×10−3 min−1

  1. Use first-order integrated rate law

For a first-order reaction, k=2.303tlog⁡([A]0[A]t)k = \frac{2.303}{t}\log\left(\frac{[A]_0}{[A]_t}\right)k=t2.303​log([A]t​[A]0​​)

Substitute the values: k=2.30360log⁡(2.40×10−21.60×10−2)k = \frac{2.303}{60}\log\left(\frac{2.40 \times 10^{-2}}{1.60 \times 10^{-2}}\right)k=602.303​log(1.60×10−22.40×10−2​)

The 10−210^{-2}10−2 terms cancel: k=2.30360log⁡(2.401.60)k = \frac{2.303}{60}\log\left(\frac{2.40}{1.60}\right)k=602.303​log(1.602.40​)

2.401.60=1.5=32\frac{2.40}{1.60} = 1.5 = \frac{3}{2}1.602.40​=1.5=23​

So, k=2.30360log⁡(32)k = \frac{2.303}{60}\log\left(\frac{3}{2}\right)k=602.303​log(23​)

  1. Evaluate the logarithm

Using the given values: log⁡3=0.477,log⁡5=0.699\log 3 = 0.477, \quad \log 5 = 0.699log3=0.477,log5=0.699

Now, log⁡2=1−log⁡5=1−0.699=0.301\log 2 = 1 - \log 5 = 1 - 0.699 = 0.301log2=1−log5=1−0.699=0.301

Hence, log⁡(32)=log⁡3−log⁡2=0.477−0.301=0.176\log\left(\frac{3}{2}\right) = \log 3 - \log 2 = 0.477 - 0.301 = 0.176log(23​)=log3−log2=0.477−0.301=0.176

  1. Calculate kkk

k=2.30360×0.176k = \frac{2.303}{60} \times 0.176k=602.303​×0.176

First, 2.303×0.176=0.4053282.303 \times 0.176 = 0.4053282.303×0.176=0.405328

Then, k=0.40532860=0.006755 min−1k = \frac{0.405328}{60} = 0.006755\ \text{min}^{-1}k=600.405328​=0.006755 min−1

Thus, k≈6.755×10−3 min−1k \approx 6.755 \times 10^{-3}\ \text{min}^{-1}k≈6.755×10−3 min−1

Nearest integer for the blank is: 777

  1. Final answer

7\boxed{7}7​

The rate constant is 7×10−3 min−1\boxed{7 \times 10^{-3}\ \text{min}^{-1}}7×10−3 min−1​

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