$\quad \mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$ In the above first order reaction, the concentration of $\mathrm{PCl}_{5}$ reduces from initial concentration $50 \mathrm{~mol} \mathrm{~L}^{-1}$ to $10 \mathrm{~mol} \mathrm{~L}^{-1}$ in 120 minutes at 300 K . The rate constant for the reaction at 300 K is $\times \times 10^{-2} \mathrm{~min}^{-1}$. the value of x is \_\_\_\_ . [Given $\log 5=0.6989$ ]

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Chemical Kinetics

PCl_{5} (g) \to PCl_{3} (g) + Cl_{2} (g)

In the above first order reaction, the concentration of

PCl_{5}

reduces from initial concentration

50 mol L^{- 1}

10 mol L^{- 1}

in 120 minutes at 300 K . The rate constant for the reaction at 300 K is

\times \times 1 0^{- 2} min^{- 1}

. the value of x is ____ .
[Given

lo g 5 = 0.6989

]

Answer: 1

A first order reaction obeys

k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ]}

. Here the concentration falls from

50

mol L

^{- 1}

10

mol L

^{- 1}

120

minutes, so

\frac{[ A ] _{0}}{[ A ]} = 5

and

k = \frac{2.303}{120} lo g 5 = \frac{2.303}{120} (0.6989) = 1.34 \times 1 0^{- 2}

min

^{- 1}

. Comparing with

x \times 1 0^{- 2}

min

^{- 1}

and rounding to the nearest integer gives

x = 1

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Integer answer