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Determining Rate Law Equations. The effect that concentration of reactants has on the rate of reaction can only be determined experimentally by analyzing patterns in data.
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Determining Rate Law Equations The effect that concentration of reactants has on the rate of reaction can only be determined experimentally by analyzing patterns in data. By keeping all conditions of the reaction constant (like temperature, pressure etc.), the overall rate of any general reaction depends only on the [reactants].
For the reaction: aA + bB cC + dD rate a [A]m[B]n • where m and n are integer values that must be experimentally determined (they do not come from the balanced chemical equation!). • m and n are known as orderof the reactant • m + n is known as the order of the reaction • rate law expression becomes rate = k[A]m[B]n by introducing the rate constant (k) and replacing the proportionality sign a. • rate constants (k) are only constant for that particular reaction under specific conditions
Eg. Determine the rate law expression for the following reaction: 2NO(g) + 2H2(g) N2(g) + 2H2O(g) Given the data: Expt. [NO] (M) [H2] (M) rate (Ms-1) 1 0.001 0.004 0.002 2 0.002 0.004 0.008 3 0.003 0.004 0.018 4 0.004 0.001 0.008 5 0.004 0.002 0.016 6 0.004 0.003 0.024
For expt 4 & 5: rate = k[NO]?[H2]? 2 = k[1][2]? ? = 1 For expt 1 & 2 rate = k[NO]?[H2]? 4 = k[2] ?[1] ? = 2 Rate law expression: rate =k[NO]2[H2]
Calculate the value of the rate constant k by rearranging the rate law expression and solving for k. Choose any expt numbers. From expt 4: k = rate[Ms-1] [NO]2[H2] [M2M] k = 0.008[s-1] (0.004)2(0.001)[M2] k = 500000 M-2s-1 k = 5 X 105 M-2s-1
Graphical Analysis of Reaction Orders Plotting concentration data vs. time ([X] vs. t) yields an exponential relationship. This graphical data is more useful when transformed into a linear relationship by plotting natural logarithm vs. time. (ln [X] vs. t)
Half-Life Half-life of a chemical reaction is defined as the time it takes for half of the chemical present to react into its products. Many applications!
Using calculus: ln[X0] = kt where Xo = initial concentration [X] X = concentration at time t ln[100] = kt1/2 (for half-life) [50] ln 2 = kt1/2 or t1/2 = 0.693/k(for 1st order reactions)