Question

For the simple decomposition reaction: AB(g) LaTeX: \longrightarrow⟶ A(g) + B(g), the rate = k{AB}2 ({ = [) and k = 0.67 1/MLaTeX: \cdot⋅s. How long (in seconds) will it take for the concentration of AB to reach one-third of its initial concentration of 2.38? Enter to 2 decimal places.

Answer 1

Answer:

$t=0.56 s]$

Explanation:

The reaction:

$AB (g) \longrightarrow A (g) + B (g)$

Reaction rate:

$r=k*[AB]^2$

$r=\frac{dC}{dt}=-k*C^2$

$dC=-k*C^2*dt$

$t=-\int_{C0}^{Cf}k*(1/C^2)*dC$

$t=-k*(1/C_0)-k*(1/C_f)$

$t=-0.67 \frac{1}{M*s}[(1/C_0)-(3/C_0)]$

$t=-0.67 \frac{1}{M*s} * (-2/2.38)$

$t=0.56 s$

Answer 2

Complete Question

.For the simple decomposition reaction: $AB_{(g)} \longrightarrow A_{(g)} + B_{(g)}$, the rate = $k[AB]^2$  and k = $0.67 1 \ L / mol \cdot s$. How long (in seconds) will it take for the concentration of AB to reach one-third of its initial concentration of $2.38\ mol /L$? Enter to 2 decimal places

Answer:

The time it would take would be    $t=1.25 sec$

Explanation:

The given decomposition reaction is

                    $AB_{(g)} -----> A_{(g)} + B_{(g)}$

Also  initial concentration of AB is     $[AB]_{o} = 2.38 mol/L$

         The rate constant is given as     $k = 0.671 \ L\ mol^{-1} \ s^{-1}$

The objective of this solution is to obtain the time(t) it would take for concentration of AB to reach one third of its initial concentration

    Now the the concentration at time t is

                      $[AB]_{t} = \frac{1}{3} [AB]_{o}$

                                $= \frac{1}{3} * 2.38$

                                $=0.7933 \ mol \ L^{-1}$f

Now this time is mathematically represented as

                 $t = \frac{\frac{1}{[AB]_t} -\frac{1}{[AB]_t} }{k}$

      substituting value

                $t = \frac{\frac{1}{0.7933} - \frac{1}{2.38} }{0.671}$

                 $t=1.25 sec$