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

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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` times
D. 8 times


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Salim Ahmed Kawsar

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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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