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-l s-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
No idea, help
You can do these by reason or mathematically. Usually the initial ones are easy enough to do by reasoning. Here is how you do it.
Look for a trial (run) in which the concn of 1 material changes and the other material does not. For example, trial 1 has 0.01 and trial 2 has 0.01 for (XO) while trial 1 shows rate of 2.5 for (O2) and trial 2 shows 5.0 for (O2).
The reasoning part is this. Forget (XO) at this point since we did not change the concn (it stayed at 0.01). We changed concn O2 from 0.01 to 0.02 (doubled it) and we changed the rate from 2.5 to 5.0 (doubled it). So doubling concn caused doubling of rate so the order is 1 because 2^what = 2; obviously what is 1.
Now we look to find a trial in which concn of O2 does not change but (XO) does.That's trial 2 and trial 3.
(XO) changes from 0.01 to 0.03 (triples) while rate changes from 5.0 to 45.0 (45.0/5.0 = 9.0 times). Tripling concn XO made rate change by 9 so 3^what = 9 and that's 2 (3 squared = 9). Therefore, the reaction is second order with respect to XO. Thus the rate law expression is
rate = k(XO)2(O2)1 and the 1 as an exponent isn't needed; I put it there to make the point that it was 1. The reaction is third order overall.
To determine the rate law for the given reaction, we need to examine how the rate of reaction changes with the concentrations of the reactants.
From the given data, we can compare the rate of reaction for different runs while keeping one reactant concentration constant and changing the other:
Comparing runs 1 and 2:
- [XO] is constant (0.010 mol L^-1) while [O2] doubles from 0.010 to 0.020 mol L^-1.
- The rate of reaction also doubles from 2.5 to 5.0 mol L^-1 s^-1.
- This indicates that the reaction rate is directly proportional to [O2].
Comparing runs 2 and 3:
- [O2] is constant (0.020 mol L^-1) while [XO] triples from 0.010 to 0.030 mol L^-1.
- The rate of reaction increases by a factor of 9, from 5.0 to 45.0 mol L^-1 s^-1.
- This shows that the reaction rate is directly proportional to the square of [XO].
Combining these observations, the rate law for the given reaction is:
rate = k[XO]^2[O2]
Therefore, the correct answer is:
a. rate = k[XO]^2[O2]
To determine the rate law for a reaction, we need to analyze the effect of changing the initial concentrations of the reactants on the rate of reaction.
Given the data provided, we can compare run #1 and run #2, while keeping the concentration of [XO] constant and doubling the concentration of [O2]. If the rate also doubles, this would suggest a first-order dependence on [O2].
Now, let's compare run #2 and run #3, while keeping the concentration of [O2] constant and tripling the concentration of [XO]. If the rate increases by a factor of 9 (3^2), this would indicate a second-order dependence on [XO].
Based on these comparisons, we can conclude that the rate law for the reaction is:
rate = k[XO]^2[O2]
Therefore, the correct answer is:
a. rate = k[XO]^2[O2]