$PC{l_5}(g) \to PC{l_3}(g) + C{l_2}(g)$
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]
Answer (integer)
1
Solution
$$PC{l_5}(g)\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{300K}^{I\,order}} PC{l_3}(g) + C{l_2}(g)$$<br><br>t = 0<br><br>50M<br><br>t = 120min<br><br>10 M<br><br>$\Rightarrow K = {{2.303} \over t}\log {{[{A_0}]} \over {[{A_t}]}}$<br><br>$\Rightarrow K = {{2.303} \over {120}}\log {{50} \over {10}}$<br><br>$\Rightarrow K = {{2.303} \over {120}} \times 0.6989 = 0.013413$ min<sup>$-$1</sup><br><br>$= 1.3413 \times {10^{ - 2}}$ min<sup>$-$1</sup><br><br>1.34 $\Rightarrow$ Nearest integer = 1
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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