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?
To determine the rate law for the given reaction, we need to analyze the data and determine the relationship between the initial concentrations of the reactants and the initial rate of reaction.
Let's compare runs 1 and 2, keeping the concentration of one reactant constant and varying the concentration of the other. In these runs, the concentration of [XO] is constant while [O2] is different.
In run 1: [XO] = 0.010 mol/L, [O2] = 0.010 mol/L, rate = 2.5 mol/(L*s)
In run 2: [XO] = 0.010 mol/L, [O2] = 0.020 mol/L, rate = 5.0 mol/(L*s)
Since only [O2] changes while [XO] is constant, we can determine the order with respect to [O2].
Changing [O2] from 0.010 to 0.020 mol/L (a factor of 2) resulted in an increase in the rate from 2.5 to 5.0 mol/(L*s). This indicates that the rate is directly proportional to the concentration of [O2]. Therefore, the order of [O2] is 1.
Next, we can compare runs 1 and 3, where [O2] is constant while [XO] is different.
In run 1: [XO] = 0.010 mol/L, [O2] = 0.010 mol/L, rate = 2.5 mol/(L*s)
In run 3: [XO] = 0.030 mol/L, [O2] = 0.020 mol/L, rate = 45.0 mol/(L*s)
Changing [XO] from 0.010 to 0.030 mol/L (a factor of 3) resulted in an increase in the rate from 2.5 to 45.0 mol/(L*s). This indicates that the rate is directly proportional to the square of [XO]. Therefore, the order of [XO] is 2.
From the above analysis, we can conclude that the rate law for the reaction is:
rate = k[XO]^2[O2]
So, the correct option is d. rate = k[XO]^2[O2]^2