Question

# The rate of a gaseous reaction is given by the expression $$k[A][B]$$. If the volume of the reaction vessel is suddenly reduced to $$1/4^{th}$$ of the initial volume, the reaction rate relating to original rate will be:

A
1/10
B
1/8
C
8
D
16

Solution

## The correct option is C $$16$$Concentration is number of moles per liter $$n/V$$. it is proportional to volume. Rate of reaction is $$k[A][B]$$ where $$[A]$$ and $$[B]$$ are a concentration of $$A$$ and $$B$$. When the volume is suddenly reduced to 1/4 times the original, the concentration changes 4 times the original concentration.The new rate will be $$k[4A][4B]=16k[A][B]$$ The new rate will be 16 times the original rate.Chemistry

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