Chemical Kinetics Question 58
Question: For the reaction, $ P+Q+R\to S $ Experimental data for the measured initial rates in given below.
Expt. (P) (Q) (R) Initial rate (in $ m{s^{-1}} $ )
- 0.2M 0.5M 0.4M $ 8.0\times {10^{-5}} $
- 0.4M 0.5M 0.4M $ 3.2\times {10^{-4}} $
- 0.4M 2.0M 0.4M $ 1.28\times {10^{-3}} $
- 0.1M 0.2M 1.6M $ 4.0\times {10^{-8}} $ The order of the reaction with respect of P, Q and R respectively is
Options:
A) 4, 1, 3
B) 2, 2, 1
C) 1, 2, 1
D) 2, 1, 1
Show Answer
Answer:
Correct Answer: D
Solution:
[d] By rate law, Rate $ =k{{[P]}^{x}}{{[Q]}^{y}}{{[R]}^{z}} $ x=order w.r.t P y=order w.r.t. Q z=order w.r.t. R
$ 8.0\times {10^{-5}}=k{{(0.2)}^{x}}{{(0.5)}^{y}}{{(0.4)}^{z}} $
$ 3.2\times {10^{-4}}=k{{(0.4)}^{x}}{{(0.5)}^{y}}{{(0.4)}^{z}} $
$ 1.28\times {10^{-3}}=k{{(0.4)}^{x}}{{(2.0)}^{y}}{{(0.4)}^{z}} $
$ 4.0\times {10^{-5}}=k{{(0.1)}^{x}}{{(0.25)}^{y}}{{(1.6)}^{z}} $
From (1) and (2),
$ \frac{3.2\times {10^{-4}}}{8.0\times {10^{-5}}}={{(2)}^{x}} $
$ (4)={{(2)}^{x}}\Rightarrow x=2 $
From (2) and (3),
$ \frac{1.28\times {10^{-3}}}{3.2\times {10^{-4}}}={{(4)}^{y}} $
$ (4)={{(4)}^{y}}\Rightarrow y=1 $
From (1) and (4),
$ \frac{8\times {10^{-5}}}{4\times {10^{-5}}}={{(2)}^{x}}{{(2)}^{y}}{{( \frac{1}{4} )}^{z}}=(4)(2){{( \frac{1}{4} )}^{z}} $
$ \Rightarrow z=1 $ Thus, order w.r.t. P=2 w.r.t. Q=1 w.r.t. R=1
