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Chemical Reaction Lab

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The rate law of a chemical reaction is a very useful equation that relates the concentration of reactants to time. The law was formed from the results of experiments by various European chemists throughout the second half of the 1800s, when they caught on that there was a correlation between the concentration of reactants and time. The most basic formula of the equation is Rate=k[A]n[B]m..., where Rate is the rate of the reaction in concentration per second (M/s), k is a constant, [A] and [B] are the concentrations of the reactants (M), and n and m are values that rate to the order of each reactant. All the values in the equation can be determined experimentally by taking the concentrations of reactants and the corresponding rate at multiple …show more content…

The first solution step is to set the rate of one trial over another to solve for one of the exponents in the rate formula. To do this trial one and two will be used: 4.69x10-4/9.38x10-4=k[0.185]n[0.133]m/k[0.185]n[0.266]m. First cancel out k, then multiply both sides of the equation by 2m to cancel out the entire right side. Next isolate 2m on the left to get 2m=2, which is equivalent to log(2)/log(2) and equals 1.00. This is the value of m. Repeat the same process for using trial one and two with m=1 and the result should be n=2.00. The values of m and n are the orders, so the reaction is second order in A and first order in B. The rate law is now Rate=k[A]2.00[B]1.00. To solve for the average value of k, the rate constant, isolate k by dividing rate by [A]2.00[B]1.00 and plug in the values for each trial. From doing this one gets four answers of k all around 0.103 (1/M2s). Add these four values and then divide by four to get the average of k=0.103 (1/M2s). Three significant figures are always used because there were three in the data and only multiplication and division were used. The final rate law is

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