Expt. | (P) | (Q) | (R) | Initial rate (in \[m{{s}^{-1}}\]) |
1. | 0.2M | 0.5M | 0.4M | \[8.0\times {{10}^{-5}}\] |
2. | 0.4M | 0.5M | 0.4M | \[3.2\times {{10}^{-4}}\] |
3. | 0.4M | 2.0M | 0.4M | \[1.28\times {{10}^{-3}}\] |
4. | 0.1M | 0.2M | 1.6M | \[4.0\times {{10}^{-8}}\] |
A) 4, 1, 3
B) 2, 2, 1
C) 1, 2, 1
D) 2, 1, 1
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), |
\[\frac{8\times {{10}^{-5}}}{4\times {{10}^{-5}}}={{(2)}^{x}}{{(2)}^{y}}{{\left( \frac{1}{4} \right)}^{z}}=(4)(2){{\left( \frac{1}{4} \right)}^{z}}\] |
\[\Rightarrow z=1\] |
Thus, order w.r.t. P=2 |
w.r.t. Q=1 |
w.r.t. R=1 |
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