The following date were obtained for the reaction
2ClO2(aq)+2OH(aq)->ClO_3^-(aq)+ClO_2^-(aq) + H2O(l)
where rate= delta[ClO_2]/ delta t
[ClO_2]0 [OH^-1]0 Initial Rate
(mol/L) (mol/L) (mol/L*s)
.0500 .100 5.75*10^-2
.100 .100 2.30*10^-1
.100 .050 1.15*10^-1
a) Determin the rate law and the value of the rate constant
b)What would be the initial rate for an expereiment with [ClO_2]0= 1.175mol/L and [OH^-1]0= .0844mol/L?
If you don't have this problem worked yet, please repost on today's board but also show what work you have done on it as well as telling us exactly what you don't understand about it.
To determine the rate law and the value of the rate constant, we can use the method of initial rates.
a) Rate law determination:
We can start by comparing the initial rates for the different experiments while keeping one reactant concentration constant.
1) Comparing experiments 1 and 2 (keeping [OH^-] constant):
Initial Rate(exp 2)/Initial Rate(exp 1) = ([ClO_2]0_2 / [ClO_2]0_1)
(2.30*10^-1 mol/L*s) / (5.75*10^-2 mol/L*s) = (.100 mol/L) / (.0500 mol/L)
4 = 2
This shows that doubling the concentration of [ClO_2]0 causes the rate to double, indicating that the reaction is first-order with respect to [ClO_2].
2) Comparing experiments 1 and 3 (keeping [ClO_2]0 constant):
Initial Rate(exp 3)/Initial Rate(exp 1) = ([OH^-]0_3 / [OH^-]0_1)
(1.15*10^-1 mol/L*s) / (5.75*10^-2 mol/L*s) = (.0500 mol/L) / (.100 mol/L)
2 = 0.5
This shows that halving the concentration of [OH^-]0 causes the rate to decrease by half, indicating that the reaction is first-order with respect to [OH^-].
Combining the results from 1) and 2), the overall rate law is:
Rate = k[ClO_2]^1[OH^-]^1
b) To determine the initial rate for an experiment with [ClO_2]0 = 1.175 mol/L and [OH^-]0 = 0.0844 mol/L, we can use the rate law equation:
Initial Rate = k[ClO_2]0[OH^-]0
Plugging in the values:
Initial Rate = k (1.175 mol/L) (0.0844 mol/L)
You would need to know the value of the rate constant (k) to determine the specific initial rate.