Question

The compound AX2 decomposes according to the equation, 2 AX2(g) => 2 AX(g) + X2(g). In one experiment, AX2 was measured at various times, and these data recorded: Time (s) {AX2} mol/L 0 0.0500 2.0 0.0448 6.0 0.0300 8.0 0.0249 10.0 0.0209 20.0 0.0088 What is the average rate from 8.0 to 20.0 seconds in mol / L s? Use 4 decimal places.

Answer 1

Answer:

0.001341 mol/L s is the average rate from 8.0 to 20.0 seconds.

Explanation:

$2 AX_2(g)\rightarrow 2 AX(g) + X_2(g)$

Average rate of the reaction =$R_a$

$R_a=-\frac{\Delta [x]}{\Delta T}=-\frac{x_2-x_1}{t_2-t_1}$

$R_a$ =Average rate of the reaction during the given time interval.

$\Delta [x]$ = Change in concentration of reactant with respect to time.

$\Delta T$ = Change in time.

$x_1$=Concentration of reactant at time$t_1$

$x_2$=Concentration of reactant at time$t_2$

So, at $t_1=8.0 sec$ the concentration of $AX_2$ :

$x_1=0.0249 mol/L$

And at $t_2=2.0 sec$ the concentration of $AX_2$ :

$x_2=0.0088 mol/L$

The average rate of the reaction at given interval will be given as:

$R_a=-\frac{x_2-x_1}{t_2-t_1}=-\frac{0.0088 mol/L-0.0249mol/L}{20.0s-8.0 s}=0.001341 mol/L s$

0.001341 mol/L s is the average rate from 8.0 to 20.0 seconds.