Question

For a different reaction, the plot of the reciprocal of concentration versus time in seconds was linear with a slope of 0.056 M-1 s -1 . If the initial concentration was 2.2 M, calculate the concentration after 100 seconds. Show your work.

Answer 1

Answer:

$C_t=0.165M$

Explanation:

From the question we are told that:

Slope $K=0.056 M-1 s -1$

initial Concentration $C_1=2.2M$

Time $t=100$

Generally the equation for Raw law is mathematically given by

$\frac{1}{C}_t=kt+\frac{1}{C}_0$

$\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0$

$C_t=0.165M$

Answer 2

The second-order reaction is the reaction that depends on the reactants of the first or the second-order reaction. The concentration after 100 seconds will be 0.165 M.

What is the specific rate constant?

The specific rate constant (k) of the second-order reaction is given in L/mol/s or per M per s. It is the proportionality constant that gives the relation between the concentration and the rate of the reaction.

Given,

Slope (k)= 0.056 per M per s

Initial concentration of the reactant $(\rm C_{1})$ = 2.2 M

Time (t) = 100 seconds

The concentration of the reaction after 100 seconds can be given by,

$\rm \dfrac{1}{C_{t}} = kt + \dfrac{1}{C_{1}}$

Substitute values in the above equation:

$\begin{aligned} \rm \dfrac{1}{C_{t}} &= 0.056 \times 100 + \dfrac{1}{2.02}\\\\&= 0.165 \;\rm M\end{aligned}$

Therefore, after 100 seconds the concentration is 0.165 M.

Learn more about the order of reaction here:

https://brainly.com/question/14810717