Question

If the volume of the container containing the gaseous mixture is increased to two times then the final rate of the reaction

Answer 1

your question is incomplete. A complete question is -> H2 + I2 -------- 2HI (an elementary reaction) Initial rate=R, if the volume of the container containing the gaseous mixture is increased 2 times, then find the rate of reaction ?

solution : rate of reaction is given by, R = k[H2][I2] = (n/V) × (n/V) =k(n/V)²

where n is number of moles and v is volume of container containing the gaseous mixture.

now a/c to question,

volume of container containing the gaseous mixture is doubled.

so, R' = k(n/V')² ,

here V' = 2V

so, R' = k(n/2V)² = k/4 (n/V)²

hence, R' = R/4

it is clear that rate of reaction becomes 1/4th of the original rate.

Answer 2

Answer:1/4 times.

Explanation:

Initial rate = K(n/v)^2

Final rate = K(n/2v)^2

= 1/4 K(n/v)^2

= 1/4 times original rate.