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Chemical Kinetics and Nuclear Chemistry question

2021 · 20 Jul · Shift 2 · Q22
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Chemical Kinetics and Nuclear Chemistry question

2021 · 20 Jul · Shift 2 · Q22

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
PCl5(g)→PCl3(g)+Cl2(g)PC{l_5}(g) \to PC{l_3}(g) + C{l_2}(g)PCl5​(g)→PCl3​(g)+Cl2​(g) In the above first order reaction the concentration of PCl5PCl_5PCl5​ reduces from initial concentration 50 mol L −-− 1 to 10 mol L −-− 1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is x ×\times× 10 −-− 2 min −-− 1. The value of x is ‾\underline{\hspace{2cm}}​. [Given log5 = 0.6989]
Numerical answer
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Correct answer: 1

  1. 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​​)

  2. Here,

  • Initial concentration: [A]0=50 mol L−1[A]_0 = 50\ \text{mol L}^{-1}[A]0​=50 mol L−1
  • Concentration after 120 min: [A]t=10 mol L−1120\ \text{min}:\ [A]_t = 10\ \text{mol L}^{-1}120 min: [A]t​=10 mol L−1
  • Time: t=120 mint = 120\ \text{min}t=120 min

So, k=2.303120log⁡(5010)k = \frac{2.303}{120}\log\left(\frac{50}{10}\right)k=1202.303​log(1050​)

  1. Simplify the ratio: 5010=5\frac{50}{10} = 51050​=5 Thus, k=2.303120log⁡5k = \frac{2.303}{120}\log 5k=1202.303​log5

  2. Using log⁡5=0.6989\log 5 = 0.6989log5=0.6989, k=2.303×0.6989120k = \frac{2.303 \times 0.6989}{120}k=1202.303×0.6989​

2.303×0.6989≈1.60962.303 \times 0.6989 \approx 1.60962.303×0.6989≈1.6096

Hence, k=1.6096120≈0.01341 min−1k = \frac{1.6096}{120} \approx 0.01341\ \text{min}^{-1}k=1201.6096​≈0.01341 min−1

  1. Write in the form x×10−2 min−1x \times 10^{-2}\ \text{min}^{-1}x×10−2 min−1: 0.01341=1.341×10−20.01341 = 1.341 \times 10^{-2}0.01341=1.341×10−2

So, x≈1.34x \approx 1.34x≈1.34

  1. Since the answer is to be filled as an integer, we take x=1x = 1x=1
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