Ammonium ion (NH4^+) reacts with nitrite ion (NO2^-) to yield nitrogen gas and liquid water. The following initial rates of reaction have been measured for the given reactant concentrations.
Expt. # [NH4^+] [NO2^-] Initial rate (M/hr)
1 0.010 0.020 0.020
2 0.015 0.020 0.030
3 0.030 0.010 0.015
What is the rate law (rate equation) for this reaction?
How much of this can you do on your own. For example, have you determined that (NH4^+) is first order? (It is.)
For order of NO2^-
rate = k(NH4^+)(NO2^-)
rate2 = k(0.015)^1(0.020)^x
rate3 = k(0.030)^1(0.010)^x
Divide rate2 equation by rate3 eqn and solve for x. I obtained x = 2.
To determine the rate law (rate equation) for this reaction, we can use the given initial rates of reaction at different reactant concentrations. The rate law can be expressed as:
Rate = k [NH4^+]^a [NO2^-]^b
where k is the rate constant, [NH4^+] is the concentration of NH4^+ ion, [NO2^-] is the concentration of NO2^- ion, and a and b are the reaction orders for NH4^+ and NO2^- respectively.
We can determine the reaction orders by comparing the initial rates of reaction for different experiments while keeping one reactant concentration constant and varying the other. Then, the reaction order of the reactant that causes a change in the initial rate when its concentration is changed can be determined by comparing the ratios of the initial rates.
Let's calculate the ratios of the initial rates for the given experiments:
Ratio 1: Rate1/Rate2
= (0.020 M/hr) / (0.030 M/hr)
= 2/3
Ratio 2: Rate2/Rate3
= (0.030 M/hr) / (0.015 M/hr)
= 2
From these ratios, we can observe that Ratio 1 is not equal to Ratio 2, indicating that the reaction orders for NH4^+ and NO2^- are not the same.
Next, let's compare the ratios of the initial rates for experiments with varying reactant concentrations to determine the reaction orders:
Ratio 3: Rate1/Rate3
= (0.020 M/hr) / (0.015 M/hr)
= 4/3
Comparing this ratio to 2^a (the exponent of [NH4^+]), it can be observed that 2^a is equal to 4/3. So, the reaction order for NH4^+ (a) is 2/3.
Similarly, comparing this ratio to 1^b (the exponent of [NO2^-]), it can be observed that 1^b is equal to 4/3. So, the reaction order for NO2^- (b) is 1.
Therefore, the rate law (rate equation) for this reaction is:
Rate = k [NH4^+]^(2/3) [NO2^-]^1
To determine the rate law for this reaction, we need to analyze the relationship between the initial rates of reaction and the concentrations of the reactants. The general form of a rate equation is:
Rate = k[A]^m [B]^n
Where [A] and [B] are the concentrations of reactants A and B, respectively, and k is the rate constant. The exponents (m and n) represent the order of the reaction with respect to each reactant.
By comparing the initial rate data provided in the table, we can determine the exponents (m and n) experimentally.
Let's consider the relationship between the initial rates in Experiments 1 and 2:
Rate1/Rate2 = (0.020 M/hr)/(0.030 M/hr) = (0.010 M)/(0.015 M)
Since the ratio of initial concentrations of NH4+ in Experiments 1 and 2 is 2:3, we can conclude that the concentration of NH4+ has a 1st order (m = 1) effect on the rate.
Now, let's consider the relationship between the initial rates in Experiments 1 and 3:
Rate1/Rate3 = (0.020 M/hr)/(0.015 M/hr) = (0.030 M)/(0.010 M)
Since the ratio of initial concentrations of NO2- in Experiments 1 and 3 is 2:3, we can conclude that the concentration of NO2- also has a 1st order (n = 1) effect on the rate.
Therefore, the rate law (rate equation) for this reaction would be:
Rate = k[NH4+]^1 [NO2-]^1
Simplifying this further, we can express the rate law as:
Rate = k[NH4+][NO2-]
So, the rate law for this reaction is first order with respect to both NH4+ and NO2-.