Question

For the reaction A+B+C→D+EA+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected:

Trial	[A][A] (MM)	[B][B] (MM)	[C][C] (MM)	Initial rate (M/sM/s)
1	0.10	0.10	0.10	3.0×10−5
2	0.10	0.10	0.30	9.0×10−5
3	0.20	0.10	0.10	1.2×10−4
4	0.20	0.20	0.10	1.2×10−4

Question 1:

Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.65 MM of reagent AA and 0.60 MM of reagents BB and CC?

Express your answer to two significant figures and include the appropriate units. Indicate the multiplication of units explicitly either with a multiplication dot or a dash.

I already asked these two questions but was given the wrong answer. Maybe whoever answers this knows the right answer. they gave me a. 14.69*10^-27 and b. 1.27*10^-4 and both are wrong

Part A:

Given the two reactions

1. H2S(aq)⇌HS−(aq)+H+(aq),  H2S(aq)⇌HS−(aq)+H+(aq),   K1K1K_1 = 9.48×10⁻⁸, and
2. HS−(aq)⇌S2−(aq)+H+(aq),  HS−(aq)⇌S2−(aq)+H+(aq),   K2K2K_2 = 1.55×10⁻¹⁹,

what is the equilibrium constant Kfinal for the following reaction?

S2−(aq)+2H+(aq)⇌H2S(aq)S2−(aq)+2H+(aq)⇌H2S(aq)

Enter your answer numerically.

Part B:

Given the two reactions

1. PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq),  PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq),   K3K3K_3 = 1.71×10⁻¹⁰, and
2. AgCl(aq)⇌Ag+(aq)+Cl−(aq),  AgCl(aq)⇌Ag+(aq)+Cl−(aq),   K4K4K_4 = 1.16×10⁻⁴,

what is the equilibrium constant Kfinal for the following reaction?

PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq)PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq)

Express your answer numerically.

Answer 1

Let the order of reaction with respect to A ,B,C is m,n,z respectively.

rate of reaction is given as

r = k[A]^m[B]ⁿ[C]^z

For trial 1

3x10^-5= k[0.1]^m[0.1]ⁿ[0.1]^z    --------- i

For trial 2

9x 10^-5= k[0.1]^m[0.1]ⁿ[0.3]^z. ---------- ii

For trial 3

1.2 x 10^-4= k[0.2]^m[0.1]ⁿ[0.1]^z.   ------------ iii

For trial 4

1.2 x 10^-4= k[0.2]^m[0.2]ⁿ[0.1]^z    -------------iv

Divide ii by i we get

3 = (3)^z

z =1

Divide iii by i

4= (2)^m

2² = 2^m

m = 2

Divide iv by iii

1 = (2)ⁿ

(2)⁰ = (2)ⁿ

n = 0

rate law for above equation is

r = k[A]²[B]⁰[C]]¹

r = k[A]²[C]

Putiing value for trial 1 in above we find value of k as

k = (3x10^-5)/(0.1)³ = 3x10^-2 M^-2s²

Now let us solve the question where the data is given as

[A] = 0.65 M

[C] 0.60 M

r = k[A]²[C]

r = 3 x 10^-2 (0.65)²(0.60) = 7.6 x 10^-3 Ms^-1