By what factor does the rate change in each of the following cases (assuming constant temperature)?
Factor (enter
as decimal if <1)
(b) A reaction is second order in reactant B, and [B] is halved?
I thought it was four but it saying it's wrong
you are sort of on the right track ... (1/2)^2 = 1/4 = .25
To determine the factor by which the rate changes when the concentration of reactant B is halved in a second-order reaction, you can use the formula for the rate equation:
Rate = k[B]^2
Where [B] represents the concentration of reactant B and k is the rate constant.
Let's consider the initial rate as R1, when [B] is at its original concentration. When [B] is halved, it becomes [B]/2.
Now, let's calculate the new rate, R2, when [B] is halved:
Rate2 = k([B]/2)^2
= (1/4)k[B]^2
= (1/4)(Rate1)
= Rate1/4
Therefore, when the concentration of reactant B is halved in a second-order reaction, the rate is reduced by a factor of 1/4, or 0.25.
To determine the factor by which the rate changes, you need to use the rate equation of the reaction. In this case, the reaction is second order in reactant B, so the rate equation can be represented as:
Rate = k[B]^2
Where [B] represents the concentration of reactant B and k is the rate constant.
Now, if [B] is halved, it means its concentration is reduced to half of its original value. Let's denote the original concentration of [B] as [B]₀ and the new concentration of [B] as [B]₁. We have:
[B]₁ = [B]₀/2
We can plug in this new concentration into the rate equation:
Rate = k([B]₁)^2
Rate = k([B]₀/2)^2
Rate = k(B₀^2/4)
Now, let's compare this new rate with the original rate. The factor by which the rate changes can be obtained by dividing the new rate by the original rate:
Factor = (Rate with [B]₁)/(Rate with [B]₀)
Factor = k(B₀^2/4) / k(B₀^2)
Factor = (1/4)
Therefore, the correct factor by which the rate changes when [B] is halved is 1/4 or 0.25, not four as you initially thought.