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For a reaction between gaseous compounds, `2A + B rarrC + D` , the reaction rate law is rate k[A][B]. If the volume of the container is made `1//4^(th)` of the initial, then what will be the rate of reaction as compared to the initial rate?A. `(1)/(16)`B. `(1)/(8)`C. `16` timesD. 8 times |
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Answer» Correct Answer - C Concentration will increase by 4 times of each reactant `therefore ` Rate will become ` (4)^(2)` I .e 16 times |
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