PCl5(g)⟶PCl3(g)+Cl2(g)In the above first order reaction, the concentration of PCl5 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 × 10-2 min−1. The value of x is …… . [Given, log 5 = 0.6989]

PCl5(g) PCl3(g)+Cl2(g) t=0 50 M t- 120 min 10M K = 2.303 t log [A0][A] K = 2.303 120 log [50][10] K= 0.013413 min-1 K= 1.34 ×10-2min-1 nearest integer= 1

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${PCl}_{5} (g) ⟶ {PCl}_{3} (g) + {Cl}_{2} (g)$
In the above first order reaction, the concentration of PCl₅ 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 × 10^-2 min⁻¹. The value of x is …… . [Given, log 5 = 0.6989]

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Detailed Solution

${PCl}_{5} (g) {PCl}_{3} (g) + {Cl}_{2} (g) t = 0 50 M t - 120 m i n 10 M K = \frac{2.303}{t} \log \frac{[A_{0}]}{[A]} K = \frac{2.303}{120} \log \frac{[50]}{[10]} K = 0.013413 m i n^{- 1} K = 1.34 \times 10^{- 2} m i n^{- 1} n e a r e s t i n t e g e r = 1$