Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided.
Q + X yields products
Trial [Q] [X] Rate
1 0.12 M 0.10 M 1.5 × 10-3 M/min
2 0.24 M 0.10 M 3.0 × 10-3 M/min
3 0.12 M 0.20 M 12.0 × 10-3 M/min
rate1 = k1*(Q1)^a*(X1)^b where a and be are the exponents for the orders and (Q) and (X1) are the concentrations for rate 1.
rate2 = k2*(Q2)^a*(X2)^b
Then
rate1......(k1*(Q1)^a*(X)^b
------------------------
rate2......(k2)*(Q2)^a*(X)^b
Pick trial 1 and 2 since (X) is the same.
1.5E-3.......(0.12)^a*(0.1)^b
-----------------------------
3.0E-3.......(0.24)^a*(0.1)^b
Since (0.1)^b in the numerator and denominator cancel,(rember k1 and k2 are equal)you have only a as the unknown and you can solve for that.
1/2 = (1/2)^a
so a must be 1.
Determine b the same way by using trials 1 and 3.
After you know a and b, go back to ANY trial, plug in (Q)and (X) and a and b and rate and solve for k.
0.012
joe beans
To determine the rate law, including the values of the orders and rate law constant, for the given reaction, we need to analyze the relationship between the concentration of reactants ([Q] and [X]) and the rate of reaction.
The rate law is generally expressed as: Rate = k[A]^m[B]^n, where Rate is the rate of the reaction, k is the rate constant, and m and n are the orders of the reactants A and B, respectively.
Let's compare the given experimental data to find the orders of Q and X.
In Trial 1, the concentration of Q is 0.12 M, the concentration of X is 0.10 M, and the rate is 1.5 × 10^-3 M/min.
In Trial 2, the concentration of Q is 0.24 M (doubled from Trial 1), the concentration of X is 0.10 M (unchanged), and the rate is 3.0 × 10^-3 M/min (also doubled from Trial 1).
Comparing Trials 1 and 2, we can see that doubling the concentration of Q leads to a doubling of the rate, indicating that Q is a first-order reactant.
In Trial 1, the concentration of Q is 0.12 M (unchanged), the concentration of X is 0.10 M (unchanged), and the rate is 1.5 × 10^-3 M/min.
In Trial 3, the concentration of Q is 0.12 M (unchanged), the concentration of X is 0.20 M (doubled from Trial 1), and the rate is 12.0 × 10^-3 M/min (eight times larger than in Trial 1).
Comparing Trials 1 and 3, we can see that doubling the concentration of X leads to an eightfold increase in the rate, indicating that X is a second-order reactant.
Therefore, the rate law for the given reaction is Rate = k[Q]^1[X]^2.
Next, let's calculate the rate law constant, k, using any of the trials. Let's use Trial 1:
Rate = k[Q]^1[X]^2
1.5 × 10^-3 M/min = k(0.12 M)^1 (0.10 M)^2
1.5 × 10^-3 M/min = k(0.012) (0.01)
1.5 × 10^-3 M/min = k(1.2 × 10^-4)
k = (1.5 × 10^-3 M/min) / (1.2 × 10^-4)
k ≈ 12.5 M^-2 min^-1
Therefore, the final rate law for the given reaction is Rate = 12.5 M^-2 min^-1 [Q]^1[X]^2, with a rate constant of 12.5 M^-2 min^-1.