consider the following reaction CHCl 3 (g) + Cl 2 (g) ----> CCl 4 (g) + HCl (g)
ID: 691717 • Letter: C
Question
consider the following reaction CHCl3 (g) +Cl2 (g) ----> CCl4 (g) + HCl (g)initial rate of the reaction was measured at several differentconcentrations of the reactants with the following results
[CHCL3] (M) [Cl2] (M) Initial rate (M/s)
.010 .010 .0035
.020 .010 .0069
.020 .020 .0098
.040 .040 .027
Determine the rate law for the reaction and the rate contant (k)for the reaction.
please show me steps.
I know the answer is k[CHCl3][Cl2]1/2
(k) = 3.5 M ^ -1/2 * s^-1
but how..
Explanation / Answer
1. Get rate for each reactant
Rate2/Rate1 = k[CHCl3]^(n/2)/k[CHCl3]^(n/1)
(0.0069M/s)/(0.0035M/s) = k(0.20M)^n/k(0.10M)^n
2 = (0.20/0.10)^n
2 = 2^n
log 2 = log (2^n)
log 2 = n log 2
n = log2/log2
n = 1
Do the same for Cl2 using rate where the concentration is changing not staying the same and you will get n = 0.5.
2. Insert rates into rate law
Rate = k [CHCl3]^1[Cl2]^0.5
= k [CHCl3][Cl2]^0.5
3. Find k, use any rate and the corresponding concentrations
k = rate/[CHCl3][Cl2]^0.5
= (0.0035 M/s)/ (0.010M)(0.010^0.5M^0.5)
= (0.0035 M/s)/ (0.010M)(0.10M^0.5)
= (0.0035 M/s) / (0.001 M^1.5)
= 3.5 M^-0.5 s^-1
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