The rate law can be determined by comparing changes in reactant concentration to changes in rate. In this reaction, doubling XY doubles the reaction rate; the reaction is first-order for XY. Doubling Z increases the rate by a factor of eight, so the reaction is third-order for Z (2³ = 8).

Knowing that Rate = k[XY] [Z]³, we can solve for k by using data from one of the trials. Plugging in data from trial 1, we can calculate the value of k.

0.16 = k[100][200]³

k = 2 * 10^-10

Let's look at each of the three reactants individually.

Compare trials 1 and 3. Doubling \(\displaystyle X\) without changing the other reactants doubles the rate from \(\displaystyle 2000\) to \(\displaystyle 4000\); therefore \(\displaystyle [X]\) has a superscript of 1.

Compare trials 3 and 4. Doubling \(\displaystyle Y\) without changing the other reactants results in a rate that is four times faster, increasing from \(\displaystyle 4000\) to \(\displaystyle 16000\). The rate increases with the square of concentration of \(\displaystyle Y\), \(\displaystyle [Y]^2\)

Compare trials 1 and 2. Doubling \(\displaystyle Z\) without changing the other reactants has no effect on the rate, which remains at \(\displaystyle 2000\), so \(\displaystyle [Z]\) is not part of the final rate equation.

The final rate law will be \(\displaystyle rate=k[X][Y]^2\).

Comparing trials 2 and 3, we see that when P_XY is constant and P_Z doubles, the rate increases by a factor of 8, or 2³. Rate is therefore proportional to [Z]³.

Comparing trials 1 and 2, we see that when P_Z is held constant and P_XY doubles, the rate doubles. Rate is therefore proportional to [XY].

Remember that in these problems, you are looking for the value of n in the function \(\displaystyle rate\ ratio=[concentration\ ratio]^n\). When doubling the concentration does not affect the reaction rate, n = 0. When doubling the concentration doubles the reaction rate, n = 1. When doubling the concentration quadruples the reaction rate, n = 2.

The order of a reaction or rate law is given by the sum of the exponents in the rate expression. Rate = k[A]¹[B]¹ is the only second-order rate law. This is the case since the reaction order is determined by the number of reactants involved. Rate = k[A]¹[B]² is not second-order; since it has one A reactant and two B reactants, this is a third-order reaction.

The coefficients in front of the reactants simply show the balanced equation, and do NOT give any information on the order of the reactants. The order of the reactants must be determined experimentally by comparing how reaction rates change when the reactant's concentration is altered.

Only the reactant in question must have an altered concentration between two trials; all others must be kept constant. When a reaction rate is doubled and the initial reaction rate is quadrupled, we determine that the reaction is second order with respect to the reactant. If the concentration of a reactant is altered but the reaction rate is unaffected, the order of the reactant is zero. Finally, if the reaction rate and reactant are multiplied by the same factor between two trials, the reactant is first order.

In the table, we can compare trial 1 and 4 to see the effect of doubling A, with constant B and C. This does not affect the rate, to A must be zeroth order. Comparing trials 2 and 3 shows that doubling B, with A and C held constant, will quadruple the rate. The reaction must be second order for B. Finally, comparing trials 1 and 2 shows us that tripling C, with A and B held constant, will triple the reaction rate. The reaction must be first order for C. This yields the rate law \(\displaystyle Rate = k[B]^{2}[C]\).

If the concentration of a reactant is doubled, \(\displaystyle 2[X]_1=[X]_2\). The relationship between this change and the change in rate is, therefore, \(\displaystyle (2[X_1])^n=[X_2]^n\), where n is the order of the reaction with respect to X. This can be reduced to \(\displaystyle (2)^n=rate_2\), allowing us to find the reaction order from the experimental rate.

\(\displaystyle (2)^0= 1 * rate_1\)

\(\displaystyle (2)^1=2*rate_1\)

\(\displaystyle (2)^2=4*rate_1\)

\(\displaystyle (2)^3=8*rate_1\)

\(\displaystyle r = k[B]^{2}[C]\)

The reaction order is equal to the sum of the exponents of the concentration variables in the rate law.

Because reactant B has an exponent of two and reactant C has an exponent of one, the total sum is three, and the reaction is therefore third order.  

To determine \(\displaystyle k\), the data from any of the three trials could be used. Using trial 1 and the rate law, we can solve for \(\displaystyle k\).

 \(\displaystyle rate = k[A][B_{2}]^{\frac{1}{2}}\)

\(\displaystyle 1.2*10^{-2} = k[1.5*10^{-3}][3.0*10^{-4}]^{\frac{1}{2}}\)

\(\displaystyle k = 460 M^{-\frac{1}{2}}s^{-1}\)

Note that the units have to be adjusted, based on the order of \(\displaystyle [A]\) and \(\displaystyle [B_2]\).

\(\displaystyle rate=\frac{M}{s}=k(M)(M)^{\frac{1}{2}}=k(M^{\frac{3}{2}})\)

\(\displaystyle k=M^{-\frac{1}{2}}s^{-1}\)

The rate law is written as \(\displaystyle Rate=k[NH_4^+]^x[NO_2^-]^y\).

Compare trials 1 and 2 to see that doubling the ammonium concentration doubles the rate. The reaction is first order for ammonium: \(\displaystyle x=1\).

Compare tirals 2 and 3 to see that tripling nitrate concentration triples the rate. The reaction is first order for nitrate: \(\displaystyle y=1\).

The final rate law is \(\displaystyle Rate=k[NH_4^+][NO_2^-]\).

The sum of the orders for each reactant gives you the overall order of the reaction. In this case, the reaction is first order with regard to each reactant, and is thus second order overall.

The rate law for the given reaction will be given by the formula:

\(\displaystyle rate=k[A]^x[B]^y\)

The variables \(\displaystyle \small x\) and \(\displaystyle \small y\) need to be determined experimentally, and the value \(\displaystyle \small k\) will be given by another formula.

\(\displaystyle k=Ae^{-\frac{E_a}{RT}}\)

The coefficient is a constant, and the variables in the exponent are the activation energy and temperature.

Adding an enzyme will lower the activation energy, affecting the rate constant. Changing the temperature will also affect the rate constant. Adjusting the ratio of reactants to products will alter the concentrations of the reactants. As the reaction runs, the rate declines as the reactants are consumed until it reaches equilibrium. Changing the reactant concentrations will change the reaction rate with respect to this equilibrium. Running the reaction in reverse will result in a completely different rate law since the reactants will be altered.

The rate law will have the general formula \(\displaystyle rate=k[A]^x[B_2]^y\). The exponents, \(\displaystyle x\) and \(\displaystyle y\), DO NOT necessarily correlate with the reaction stoichiometry. They must be determined from the experimental data.

To determine \(\displaystyle x\), compare trials 1 and 3, where \(\displaystyle [B_2]\) is held constant.

\(\displaystyle \frac{rate 3}{rate 1} = \frac{k[A]^{x}[B_{2}]^{y}}{k[A]^{x}[B_{2}]^{y}}\)  

Since \(\displaystyle k\) is constant and \(\displaystyle [B_2]\) is the same for both experiments, by plugging in the values for rate and \(\displaystyle [A]\) for trials 1 and 3, we get this equation.

\(\displaystyle \frac{(2.4*10^{-2})}{(1.2*10^{-2})} = 2 = (\frac{3.0 * 10^{-3}}{1.5*10^{-3}})^{x}\).  

This reduces to \(\displaystyle 2=2^x\), which means \(\displaystyle x=1\).  

To find \(\displaystyle y\), compare trials 1 and 2, where \(\displaystyle [A]\) is held constant. Following the same type of calculation as shown above would give \(\displaystyle y=\frac{1}{2}\).

\(\displaystyle \frac{rate2}{rate1}=\frac{k[A]^x[B_2]^y}{k[A]^x[B_2]^y}\)

\(\displaystyle \frac{(2.4*10^{-2})}{(1.2*10^{-2})} = 2 = (\frac{1.2*10^{-3}}{3.0*10^{-4}})^{y}\)

\(\displaystyle 2=(4)^y\)

\(\displaystyle y=\frac{1}{2}\)

The rate law is \(\displaystyle rate = k[A][B_{2}]^{\frac{1}{2}}\).