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Reaction Rate Law Lab Report

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Review 2: Text In chemical kinetics, the reaction rate law is calculated experimentally to find how change in concentration affects change in rate, and to find a proportionality constant k, known as the rate constant. Since the instantaneous initial rates at various concentrations of A and B are provided, we must find the orders m and n of each reagent to determine the rate law, Rate=k[A]m[B]n. When A is doubled and B is held constant, the instantaneous initial rate approximately quadruples. Between Trials 1 and 3 where [B]=0.133 M, the change in [A} is .370 M/.185 M=2.00, and instantaneous initial rate changes by 1.88*10-3/4.69*10-4=4.01≅4. Between Trials 2 and 4 where [B]=0.266 M, [A] changes by .370/.185 and instantaneous initial rate changes by 3.75*10-3/9.38*10-4=4.00. Therefore since the change in rate is the change in [A] squared, A is a second order reagent. When B is doubled and A is held constant, the instantaneous initial rate approximately doubles. Between Trials 1 and 2 where [A]=0.185 M, the change in [B] is .266 …show more content…

Since rate=-d[A]/dt and k[A]^2=-d[A]/dt, rearranging the terms yields k*dt=-d[A]/[A]^2. Integrating from At to Ao with respect to A, and t to 0 with respect to t yields -1/[A]t+1/[A]o=kt. In the form of y=mx+b, 1/[A]t=1/[A]o+kt where slope=k. The appropriate linear graph for B is between ln[B] and time.Since k[B]=-d[B]/dt, this rearranges to k*dt=-d[B]/[B]. Integrating the left side from t to 0 and the right from Bt to Bo yields kt=-ln[B]t+ln[B}o. This equals ln[B]t=-kt+ln[B]o where slope=-k.. Experimentally, a pseudo approximation would be used to determine k by assuming an excess of the constant reagent. In this case, Rate=k’[A]^2 and k=k’[B] if B is constant, and vice versa. The determined values for order and k can then be substituted into the overall rate

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