JEE MAIN - Chemistry (2021 - 20th July Evening Shift - No. 22)
$$PC{l_5}(g) \to PC{l_3}(g) + C{l_2}(g)$$
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 $$\times$$ 10$$-$$2 min$$-$$1. The value of x is __________. [Given log5 = 0.6989]
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 $$\times$$ 10$$-$$2 min$$-$$1. The value of x is __________. [Given log5 = 0.6989]
Answer
1
Explanation
$$PC{l_5}(g)\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{300K}^{I\,order}} PC{l_3}(g) + C{l_2}(g)$$
t = 0
50M
t = 120min
10 M
$$ \Rightarrow K = {{2.303} \over t}\log {{[{A_0}]} \over {[{A_t}]}}$$
$$ \Rightarrow K = {{2.303} \over {120}}\log {{50} \over {10}}$$
$$ \Rightarrow K = {{2.303} \over {120}} \times 0.6989 = 0.013413$$ min$$-$$1
$$ = 1.3413 \times {10^{ - 2}}$$ min$$-$$1
1.34 $$\Rightarrow$$ Nearest integer = 1
t = 0
50M
t = 120min
10 M
$$ \Rightarrow K = {{2.303} \over t}\log {{[{A_0}]} \over {[{A_t}]}}$$
$$ \Rightarrow K = {{2.303} \over {120}}\log {{50} \over {10}}$$
$$ \Rightarrow K = {{2.303} \over {120}} \times 0.6989 = 0.013413$$ min$$-$$1
$$ = 1.3413 \times {10^{ - 2}}$$ min$$-$$1
1.34 $$\Rightarrow$$ Nearest integer = 1
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