| \[[A](mol\,{{L}^{-1}})\] | \[[B](mol\,{{L}^{-1}})\] | Initial Rate \[(mol\,{{L}^{-1}}{{s}^{-1}})\] |
| 0.05 | 0.05 | 0.045 |
| 0.10 | 0.05 | 0.090 |
| 0.20 | 0.10 | 0.72 |
A) \[Rate=k\left[ A \right]\left[ B \right]\]
B) \[Rate=k{{\left[ A \right]}^{2}}{{\left[ B \right]}^{2}}\]
C) \[Rate=k\left[ A \right]{{\left[ B \right]}^{2}}\]
D) \[Rate=k{{\left[ A \right]}^{2}}\left[ B \right]\]
Correct Answer: C
Solution :
\[r=K{{\left[ A \right]}^{x}}{{\left[ B \right]}^{y}}\] \[0.045=K{{\left( 0.05 \right)}^{x}}{{\left( 0.05 \right)}^{y}}\] ....(a) \[0.090=K{{\left( 0.10 \right)}^{x}}{{\left( 0.05 \right)}^{y}}\] ....(b) \[0.72=K{{\left( 0.20 \right)}^{x}}{{\left( 0.10 \right)}^{y}}\] ....(c) From (a)\[\div \](b), \[\frac{0.045}{0.090}={{\left( \frac{0.05}{0.10} \right)}^{x}}\Rightarrow x=1\] From (b)\[\div \](c), \[\frac{0.090}{0.720}={{\left( \frac{0.10}{0.20} \right)}^{x}}.{{\left( \frac{0.05}{0.10} \right)}^{y}}\Rightarrow y=2\] Hence, \[r=K\left[ A \right]\text{ }{{\left[ B \right]}^{2}}\] Correct option : (c)
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