A reaction in which A, B, and C react to form products is zero order in A, one-half order in B, and second order in C. If the concentration of C is increased by a factor of 4.95 (and all other concentrations are held constant), how much will the rate of reaction increase?
Won't that be (4.95)^2? or did I miss something in the problem?
To determine how the rate of reaction will change when the concentration of C is increased by a factor of 4.95, we need to analyze the reaction order with respect to C.
The rate equation for this reaction can be written as follows:
Rate = k[A]^0[B]^0[C]^2
Since the reaction is zero order in A and one-half order in B, both [A] and [B] are raised to the power of zero in the rate equation, meaning their concentrations do not affect the rate of reaction. Therefore, any changes in their concentrations will not have an impact on the rate.
However, the reaction is second order in C, which means the rate is directly proportional to the square of its concentration.
Now, let's denote the initial concentration of C as [C]₀, and the increased concentration as [C]₁.
According to the question, [C]₁ = 4.95[C]₀.
Using this information, we can calculate how the rate of reaction changes:
Rate₁ = k[A]^0[B]^0[C]₁^2
= k(1)(1)(4.95[C]₀)^2
= 24.5025k[C]₀^2
Rate₀ = k[A]^0[B]^0[C]₀^2
= k(1)(1)([C]₀)^2
= k[C]₀^2
To find the change in rate, we can take the ratio of Rate₁ to Rate₀:
Change in rate = (Rate₁ - Rate₀) / Rate₀
Substituting the values:
Change in rate = (24.5025k[C]₀^2 - k[C]₀^2) / k[C]₀^2
= 23.5025k[C]₀^2 / k[C]₀^2
= 23.5025
Therefore, the rate of reaction will increase by a factor of 23.5025 when the concentration of C is increased by a factor of 4.95 (and all other concentrations are held constant).