The gas phase reaction X+Y⇄ Z is first order in Y and second order in X. If the concentration of X is double and concentration of Y is tripled. The reaction rate would be increase by a factor of:
then,
Rate= k [x]^2[y] --> k[x]^4[y}^3
is this the correct line of thought? from this rate law how would one determine the amount that the rate is increased?
The right line of thought but you missed a turn somewhere.
That's rate = k[x]^2[y]^1, then
rate = k(2)^2(3)^1 = 12*k
rate=k [x]^2[y]
if you double x, then rate factor goes up by 4, and triple y, up by 3
overall, the rate has increased by 12
No your line of thought is nor correct.
Yes, the correct rate law for the given gas phase reaction is Rate = k[X]^2[Y].
To determine how the rate changes when the concentration of X is doubled and the concentration of Y is tripled, we can use the rate law:
Rate = k[X]^2[Y]
Let's denote the initial rate as R1 and the final rate after the concentration changes as R2.
Initial rate (R1) = k[X]₁^2[Y]₁
The concentration of X is doubled, so [X]₂ = 2[X]₁, and the concentration of Y is tripled, so [Y]₂ = 3[Y]₁.
Final rate (R2) = k[X]₂^2[Y]₂
= k(2[X]₁)^2(3[Y]₁)
= 12k[X]₁^2[Y]₁
To determine the amount that the rate is increased, we can compare the final rate (R2) to the initial rate (R1):
(R2 - R1)/R1 * 100%
= [(12k[X]₁^2[Y]₁) - (k[X]₁^2[Y]₁)] / (k[X]₁^2[Y]₁) * 100%
= (11k[X]₁^2[Y]₁) / (k[X]₁^2[Y]₁) * 100%
= 11 * 100%
= 1100%
Therefore, the rate would be increased by a factor of 1100% or 11-fold when the concentration of X is doubled and the concentration of Y is tripled.