A student performs the reaction below in three experiments studying initial concentrations and initial rates. The data is summarised in the table below. Show all your work.
2 NO(g) + Br2(g) → 2 NOBr(g)
a. What is the order of this reaction with respect to NO? ____
b. What is the order of this reaction with respect to Br2? ____
c. What is the overall order of this reaction? ____
d. What is the rate law constant, k, for this reaction? (Be sure to include the value and the units.)
[NO] [Br2] Rate (mol/L·s)
0.80 0.50 0.17
1.60 0.50 0.34
0.80 1.00 0.68
rate = k(NO)^x (Br2)^y where x and y are the orders for NO and Br2 respectively. You want to look in the table and find two concentrations that are the same. For (NO) I see trials 1 and 3. So that part looks this way.
rate 3 = k3(NO)^x(Br2)^y
----------------------------------(this is a fraction. Top numerator;lower denomin
rate 1 = k1(NO)^x(Br2)^y
Substitute 0.80 for NO in numerator and denominator, Substitute 1.00 for (Br2) for numerator and 0.50 for (Br2) in denominator. Substitute the rates also. It will look like this.
rate 3 = k3(NO)^x(Br2)^y
-----------------------------------
rate 1 = k1(NO)^x(Br2)^y
0.68 = k3(0.80)^x(1.00)^y
-------------------------------------
0.17 = k1(0.80)^x(0.50)^y
You see (0.80)^x cancels. 0.68/0.17 = 4 while (1.00/0.5)^y = 2^y
This is the same reaction so k3 = k1 and that cancels and you're left with
4 = 2^y so that tells you y must be 2; therefore, the reaction is second order with respect to Br2. You can see how difficult this typing and fractions is so I won't do NO but you follow the same procdure to do that part.
c. Overall order is order for NO + order for Br2 = 1 + 2 = ?
d. To determine the rate law constant k, take ANY trial, substitute the values for NO, Br2, exponents for x and y and rate into the generic rate law equation and calculate k. The equation to use is:
rate = k(NO)^x (Br2)^y where x and y are the orders for NO and Br2
Post your work if you get stuck.
To determine the order of the reaction with respect to NO and Br2, we need to compare the effect of changing the initial concentration of each reactant on the reaction rate. We can use the method of initial rates to do this.
The general rate law equation for a reaction is:
rate = k[NO]^x[Br2]^y
where k is the rate constant, [NO] is the concentration of NO, [Br2] is the concentration of Br2, and x and y are the orders of the reaction with respect to NO and Br2, respectively.
To determine the order of the reaction with respect to NO, we compare experiments 1 and 3. In these experiments, the initial concentration of NO is the same (0.80 mol/L), while the concentration of Br2 is different. Since the rate doubles when [Br2] doubles, we can conclude that the order of the reaction with respect to NO is 1.
To determine the order of the reaction with respect to Br2, we compare experiments 1 and 2. In these experiments, the initial concentration of Br2 is the same (0.50 mol/L), while the concentration of NO is different. Since the rate does not change when [NO] changes, we can conclude that the order of the reaction with respect to Br2 is 0.
Now, to determine the overall order of the reaction, we simply add the individual orders together:
overall order = order with respect to NO + order with respect to Br2
= 1 + 0
= 1
Therefore, the overall order of the reaction is 1.
Finally, to calculate the rate law constant k, we can choose any of the experiments and substitute the values of [NO], [Br2], and the rate into the rate law equation:
rate = k[NO]^x[Br2]^y
By plugging in the values from experiment 1 (0.80 mol/L for [NO], 0.50 mol/L for [Br2], and 0.17 mol/L·s for the rate), we can solve for k:
0.17 mol/L·s = k(0.80 mol/L)^1(0.50 mol/L)^0
k = 0.17 mol/L·s / (0.80 mol/L)
k = 0.2125 s^-1
Therefore, the rate law constant k for this reaction is 0.2125 s^-1.