Question

Initial rate data are listed in the table for the reaction:

Initial rate data are listed in the table for the reaction: NH4+ (aq) + NO2 (aq) + N2 (g) + H20 (1) Experiment (NH4+]: [NO 2-1. Initial rate (M/s) 0.24 0.10 7.2 x 10-4 0.10 3.6 x 10-4 10.12 0.15 5.4 x 10-4 0.12 0.12 4.3 x 10-4 0.12 First determine the rate law and rate constant. Under the same initial conditions as in Experiment 4, calculate [NH4+] at 184 seconds after the start of the reaction. In this experiment, both reactants are present at the same initial concentration. The units should be M, and should be calculated to three significant figures. Submit Answer Tries 0/45

Answer 1

let rate law is

Rate = K [NH₄⁺]^x [NO₂^-]^y

now,

Expt 1.

7.2 *10^-4 = K * ( 0.24)^x * ( 0.10)^y (1)

Expt 2

3.6*10^-4 = (0.12)^x *( 0.10)^y   (2)

Expt 3

5.4 *10^-4 = (0.12)^x * (0.15)^y (3)

Expt 4

4.3*10^-4 = (0.12)^x *. (0.12)^y (4)

Now,

Now. Eq. 1 / Eq. 2

7.2 *10^-4 = K * ( 0.24)^x * ( 0.10)^y (1)

Expt 2

$\small \frac{7.2*10^{-4}}{3.6*10^{-4}}$ =  $\small \frac{K * (0.24)^{x} (0.10)^{y}}{K *(0.12)^{x}(0.10)^{y}}$

or,

2 = (2)^x

or, x   = 1

again

Eq. 3 / Eq. 2

$\small \frac{5.4*10^{-4}}{3.6*10^{-4}}$ =   $\small \frac{K * (0.12)^{x} (0.15)^{y}}{K *(0.12)^{x}(0.10)^{y}}$

or, 1.5 = (1.5)^y

or, y = 1

therefore rate law equation is

Rate = K * [NH₄⁺]¹ [NO₂^-]¹

overall order = 1+ 1= 2

using the values of Expt 1

7.2 *10^-4 = K * ( 0.24) * ( 0.10)

or, K = 7.2 *10^-4 / (0.24*0.10)

or, K = 0.03 M ^-1 S^-1

From the balanced reaction ,

rate of reaction = -  $\small \frac{\mathrm{d} [NH_{4}^{+}]}{\mathrm{d} t}$ =  K * [NH₄⁺] [NO₂^-]

given, dt = 184 sec

[NH₄⁺] = 0.12 M

[NO₂^-] = 0.12 M

K = 0.03 M ^-1 S^-1

so,

$\small \frac{\mathrm{d} [NH_{4}^{+}]}{184}$ = 0.03 * 0.12* 0.12  

or, $\small \frac{\mathrm{d} [NH_{4}^{+}]}{184}$ = 0.00043

or, d [NH₄⁺] = 0.079 M

or,   [NH₄⁺] after 184 sec after start of the reaction

= 0.12 - 0.079  = 0.041 M