Ineed with a, b, c, d, and e on the long answer

Answers:

(a) Second order in A

(b) Second order in B

(c)     r = k[A]²[B]²

(d)    k = 0.0123 L³mol⁻³s⁻¹

(e) [A] = 8.86 mol·L⁻¹

Step-by-step explanation:

A + B ⟶ C

The rate law is r = k [A]^m[b]ⁿ

Our problem is to determine the values of m and n.

We use the method of initial rates to determine the order of reaction with respect to a component.

(a) Order with respect to A

We must find a pair of experiments in which [A] changes, but [B] doesn't.

They would be Experiments 3 and 1.

[B] is constant, so only [A] is changing the rate

r₃/r₁ = {k[A]₃^m}/{k[A]₁^m }     Cancel the ks

r₃/r₁ = {[A]₃/[A]₁}^m

80/5 = (40.2/10.1)^m

   16 = 3.98^m

(i) Method 1: by inspection

16 ≈4²

If quadrupling the concentration multiplies the rate by a factor of 16 (4²), the reaction is 2nd order.

(ii) Method 2: mathematical

    16 = 3.98^m

log16 = mlog3.98     Take the logarithm of each side

    m = (log16)/(log3.98)

       = 1.204/0.5999

       = 2.007             Round off to nearest integer

       ≈ 2

By either method, the reaction is second order in A.

r = k[A]²

(b) Order with respect to B

We must find a pair of experiments in which [B] changes, but [A] doesn't. There are none.

However, we know the effect of A on the rate.

Choose a different pair, say, Experiments 2 and 1.

       r₂/r₁ = ([A₂]/[A]₁)²([B]₂/[B₁]^m

     80/5 = (19.8/10.1)²(3.99/2.01)^m

         16 = (1.960)²(1.985)ⁿ

         16 = 3.843 × 1.985ⁿ     Divide each side by 3.843

    4.163 = 1.85ⁿ                     Take the logarithm of each side

log4.163 = nlog1.985            Divide each side by log1.985

           n = log4.163/log1.985

              = 2.08

              ≈ 2

The reaction is second order in B.  

rate = k[B]²

(c) Overall rate law

The reaction is second order in A and second order in B.

The overall rate law for the reaction is

r = k[A]²[B]²

(d) Value of k

Choose any experiment (say, Experiment 3) and insert the known values.

                    r = k[A]²[B]²

80 mol·L⁻¹·s⁻¹ = k(40.2 mol·L⁻¹)²(2.00 mol·L⁻¹)²

                      = k × 1616 mol²L⁻² × 4.00 mol²L⁻²

                      = k × 6464 mol⁴L⁻⁴     Divide each side by 6464 mol⁴L⁻⁴

                  k = (80 mol·L⁻¹s⁻¹)/(6464 mol⁴L⁻⁴)

                     = 0.0123 L³mol⁻³s⁻¹

(e) [A] in Experiment 4

    r = k[A]²[B]²

35 =  0.0123 × [A]²[6.00]²

      = 0.4456[A]²     Divide each side by 0.4456

[A]² = 35/0.4456

      = 78.6               Take the square root of each side

[A] = 8.86 mol·L⁻¹

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gas

explanation:

smoke is the matter of a gas