(a) What is the rate law for this reaction? Rate = k [HgCl 2 (aq)] [C 2 O 4 2- (

Question

(a) What is the rate law for this reaction?

Rate = k [HgCl2(aq)] [C2O42-(aq)]Rate = k [HgCl2(aq)]2 [C2O42-(aq)]    Rate = k [HgCl2(aq)] [C2O42-(aq)]2Rate = k [HgCl2(aq)]2 [C2O42-(aq)]2Rate = k [HgCl2(aq)] [C2O42-(aq)]3Rate = k [HgCl2(aq)]4 [C2O42-(aq)]

(b) What is the value of the rate constant?





(c) What is the reaction rate when the concentration of HgCl2(aq) is 0.223 M and that of C2O42-(aq) is 0.301 M if the temperature is the same as that used to obtain the data shown above?

M/s

Experiment [HgCl2(aq)] (M) [C2O42-(aq)] (M) Rate (M/s) 1 0.147 0.147 3.62e-05 2 0.147 0.294 0.000145 3 0.294 0.147 7.24e-05 4 0.294 0.294 0.000290

Explanation / Answer

(a) let the rate law be,

rate = k[HgCl2]^x[C2O4^2-]^y

with, x and y be the order with respect to HgCl2 and C2O4^2- respectively

k be the rate constant

compare Expt. 1 and 2,

conc. of HgCl2 is same and gets cancelled out

rate1/rate2 = 3.62 x 10^-5/0.000145 = (0.147/0.294)^y

Take log on both sides,

y = 2

So the reaction is second order with respect to C2O4^2-

Compare Expt. 2 and 4,

conc. of C2O4^2- is same and gets cancelled out

rate 2/rate 4 = 0.000145/0.000290 = (0.147/0.294)^x

take log on both sides,

x = 1

So reaction is first order with respect to HgCl2

the rate law thus becomes,

rate = k[HgCl2][C2O4^2-]^2

(b) Rate constant

take expt. 2

k = 0.000145/(0.147)(0.294)^2 = 0.0114 M-2.s-1

(c) When [HgCl2] = 0.223 M and [C2O4^2-] = 0.301 M

we have,

rate = 0.0114(0.223)(0.301)^2 = 2.303 x 10^-4 M/s

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