Medium
For a gaseous reaction $2A + B \rightarrow C + D$,the rate of reaction is given by $Rate = K[A][B]$. If the volume of the container is reduced to $1/4$ of its original volume,what will be the ratio of the new rate to the original rate?
- A$16$
- B$4$
- C$1/16$
- D$1/8$
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$Exp.$ $[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ Initial rate $(mol \ L^{-1} \ s^{-1})$ $1$ $2.5 \times 10^{-4}$ $3 \times 10^{-5}$ $5 \times 10^{-4}$ $2$ $5 \times 10^{-4}$ $6 \times 10^{-5}$ $4 \times 10^{-3}$ $3$ $1 \times 10^{-3}$ $6 \times 10^{-5}$ $1.6 \times 10^{-2}$
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| $1$ | $2.5 \times 10^{-4}$ | $3 \times 10^{-5}$ | $5 \times 10^{-4}$ |
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Medium
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