| [A] | [B] | [C] | rate(M/sec.) |
| 0.2 | 0.1 | 0.02 | \[8.08\times {{10}^{-3}}\] |
| 0.1 | 0.2 | 0.02 | \[2.01\times {{10}^{-3}}\] |
| 0.1 | 1.8 | 0.18 | \[6.03\times {{10}^{-3}}\] |
| 0.2 | 0.1 | 0.08 | \[6.464\times {{10}^{-2}}\] |
A) \[-1,\]\[1,\]\[3/2\]
B) \[-1,\]\[1,\]\[1/2\]
C) \[1,\]\[3/2,\]\[-1\]
D) \[1,\]\[-1,\]\[3/2\]
Correct Answer: D
Solution :
[d] If rate\[=k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}}\] From first two given data \[8.08\times {{10}^{-3}}=k{{[0.2]}^{x}}{{[0.1]}^{y}}{{[0.02]}^{z}}\] ....(1) \[2.01\times {{10}^{-3}}=k{{[0.1]}^{x}}{{[0.2]}^{y}}{{[0.02]}^{z}}\] ....(2) Divide (1) by (2) we get, \[4={{2}^{x}}{{(1/2)}^{y}}\] Similarly, from second and third data \[{{(9)}^{y}}{{(9)}^{z}}=3\] \[2y+2z=1\] From first and fourth data \[{{4}^{z}}=8={{2}^{3}}\] \[2z=3\]. So \[z=3\text{/}2,\,\,y=-1,\,\,x=1\]
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