For the reaction, 2 XO + O2 = 2 X02, some data obtained from measurement of the initial rate of reaction at varying concentrations are given below.
run # [XO] [O2] rate, mol L-ls-l
1 0.010 0.010 2.5
2 0.010 0.020 5.0
3 0.030 0.020 45.0
The rate law is therefore
a. rate = k[XO]2 [O2] b.rate = k[XO][O2]2 c.rate = k[XO][O2]
d. rate = k[XO]2 [O2] 2 e.rate = k[XO]2 / [O2] 2
I chose rate=k(XO)^2 (O2)^2. Is this correct?
No. It is 2nd order with respect to XO but third order over all (which should tell you the exponent for O2.
Would it be k[XO]^2[O2] because second order is squared?
Yes.
To determine the rate law for the reaction, we can examine the data provided and use the method of initial rates.
The rate law represents the relationship between the concentrations of reactants and the rate of the reaction. Using the given data, we compare the rates of the different runs and see how they change with respect to the concentrations.
Let's analyze the data:
Run # [XO] [O2] rate (mol L^-1 s^-1)
1 0.010 0.010 2.5
2 0.010 0.020 5.0
3 0.030 0.020 45.0
By comparing runs 1 and 2, we can see that when [XO] is constant (0.010 M) and [O2] doubles (0.010 M to 0.020 M), the rate also doubles (2.5 to 5.0). This suggests that the rate is directly proportional to the concentration of [O2], but not [XO].
By comparing runs 2 and 3, we observe that when [XO] triples (0.010 M to 0.030 M) and [O2] remains constant (0.020 M), the rate increases by 9 times (5.0 to 45.0). This indicates that the rate is directly proportional to the square of [XO], but not [O2].
Therefore, based on the given data, the rate law can be determined as:
rate = k[XO]^2[O2]
Thus, your choice of rate = k[XO]^2[O2]^2 is not correct. The correct rate law is rate = k[XO]^2[O2].