| \[A/mol\,{{L}^{-1}}\] | 0.20 | 0.20 | 0.40 |
| \[B/mol\,{{L}^{-1}}\] | 0.30 | 0.10 | 0.05 |
| \[{{r}_{0}}/\]\[mol\,{{L}^{-1}}{{s}^{-1}}\] | \[5.07\times {{10}^{-5}}\] | \[5.07\times {{10}^{-5}}\] | \[1.43\times {{10}^{-4}}\] |
A) 2.5, 1.0
B) 1.5,0
C) 2.5,0
D) 1.5, 1
Correct Answer: B
Solution :
Rate\[=k{{[A]}^{p}}{{[B]}^{q}}\] \[Rat{{e}_{1}}={{(0.20)}^{p}}{{(0.30)}^{q}}=5.07\times {{10}^{-5}}\] ... (i) \[Rat{{e}_{2}}={{(0.20)}^{p}}{{(0.10)}^{q}}=5.07\times {{10}^{-5}}\] ... (ii) \[\frac{{{(0.30)}^{q}}}{{{(0.10)}^{q}}}=1\] \[{{3}^{q}}=1\]or\[{{3}^{q}}={{3}^{0}}\]\[\therefore \]\[q=0\] \[Rat{{e}_{3}}={{(0.40)}^{p}}{{(0.05)}^{q}}=1.43\times {{10}^{-4}}\] ? (iii) Dividing (iii) by (i), we get \[\frac{{{(0.40)}^{p}}{{(0.05)}^{q}}}{{{(0.20)}^{p}}{{(0.10)}^{q}}}=\frac{1.43\times {{10}^{-4}}}{5.07\times {{10}^{-5}}}\] \[{{2}^{p}}\times \frac{1}{{{2}^{q}}}=2.8\] Since, \[q=0\] \[{{2}^{p}}\times \frac{1}{{{2}^{0}}}=2.8\] \[{{2}^{p}}=2.8\] \[{{2}^{p}}={{(2)}^{1.5}}\] \[\therefore \] \[p=1.5\] Thus, order of reaction w.r.t.\[A=1.5,\]and w.r.t. \[B=0\].
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