Question

Given that the rate constant for the decomposition of hypothetical compound X from part A is 1.60 M−1⋅min−1, calculate the concentration of X after 13.0min .
(min) [X](M)
0 0.467
1 0.267
2 0.187
3 0.144
4 0.117
5 0.099
6 0.085
7 0.075

Answer 1

The concentration of X after $13{\text{ minutes}}$ is $\boxed{0.04359}.$

Further explanation:

Given:

The rate constant for the decomposition $k = 1.6{M^{ - 1}}{m^{ - 1}}.$

Explanation:

The second order integrated rate law can be expressed as follows,

$\boxed{\frac{1}{{{N_0}}} = \frac{1}{{{N_1}}} - kt}$

Here, ${N_0}$ represents the initial concentration, ${N_1}$ represents the concentration after time t and k represents the constant.

The concentration after 13 minutes can be obtained as follows,

$\begin{aligned}\frac{1}{{{N_{13}}}} - kt &= \frac{1}{{{N_0}}}\\ \frac{1}{{{N_{13}}}} - 1.6\left( {13} \right) &= \frac{1}{{0.467}}\\\frac{1}{{{N_{13}}}} &= 2.14 + 20.8\\\frac{1}{{{N_{13}}}} &= 22.94\\{N_{13}} &= \frac{1}{{22.94}}\\\end{aligned}$

The concentration of X after $13{\text{ minutes}}$ is $\boxed{0.04359}.$

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Exponential function

Keywords: rate, constant, decomposition, integral, hypothetical, compound, X, part A, 1.60, concentration, 13 minutes.

Answer 2

Using the 2nd order integrated rate law.

1/A1 = 1/A0  + Kt 

A1 = 0.267
A0 = 0.467
k = 1.6
t = time

1/A13 = 1/0.467 + 1.6(13)
1/A13 = 1/0.467 +  20.8
1/A13 = 2.141 + 20.8
1/A13 = 22.941
1 = 22.941 * A13
1/22.941 = A13
A13 = 0.04359

The concentration of X after 13 minutes is 0.04359