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.
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
1 answer