A reaction in which X and Y form products was found to be second order in X and first order in Y. By what factor will the rate of the reaction change if [X] is doubled and [Y] is tripled?
a. 10
b. 12
c. 20
d. 2
2X + Y = X2Y
rate = (X)^2(Y)
What if (X) = 1 and (Y) = 1
then rate = (1)^2(1) = 1
Now we double X and triple y so
rate = (2)^2(3) = 12
So you have changed it from rate of 1 to rate of 12. What factor is that.
Oh I see, so it's 12. Thank you!!
To determine how the rate of the reaction changes when [X] is doubled and [Y] is tripled, we need to consider the relationship between the rate of a reaction and the concentrations of the reactants.
According to the given information, the reaction is second order in X and first order in Y. This means that the rate equation for the reaction can be expressed as:
rate = k[X]^2[Y]
Where k is the rate constant.
Now, let's consider the initial rate of the reaction (Rate1) when [X] and [Y] are at their original concentrations.
Rate1 = k[X₁]^2[Y₁] ...(1)
Next, let's determine the rate of the reaction (Rate2) when [X] is doubled and [Y] is tripled.
Rate2 = k[2X₁]^2[3Y₁] = k(2^2)(3)[X₁]^2[Y₁]
= 12k[X₁]^2[Y₁] ...(2)
To find the factor by which the rate of the reaction changes, we can take the ratio of Rate2 to Rate1:
Rate2 / Rate1 = (12k[X₁]^2[Y₁]) / (k[X₁]^2[Y₁]) = 12
Therefore, the rate of the reaction will change by a factor of 12 when [X] is doubled and [Y] is tripled.
Hence, the correct answer is b. 12.
To determine the factor by which the rate of the reaction will change, we need to apply the rate equation.
From the information given, we know that the reaction is second order in X and first order in Y. This implies that the rate equation can be expressed as:
Rate = k[X]^2[Y]
Where [X] and [Y] represent the concentrations of X and Y, respectively, and k is the rate constant.
Now, let's consider how the rate of the reaction changes when [X] is doubled and [Y] is tripled:
Doubling [X] means that [X] becomes 2[X].
Tripling [Y] means that [Y] becomes 3[Y].
Substituting these new concentrations into the rate equation, we have:
Rate' = k(2[X])^2(3[Y]) = k(4[X]^2)(3[Y]) = 12k[X]^2[Y]
So, the rate of the reaction is multiplied by a factor of 12 when [X] is doubled and [Y] is tripled.
Therefore, the correct answer is b. 12.