Answer:
0.001341 mol/L s is the average rate from 8.0 to 20.0 seconds.
Explanation:
[tex]2 AX_2(g)\rightarrow 2 AX(g) + X_2(g)[/tex]
Average rate of the reaction =[tex]R_a[/tex]
[tex]R_a=-\frac{\Delta [x]}{\Delta T}=-\frac{x_2-x_1}{t_2-t_1}[/tex]
[tex]R_a[/tex] =Average rate of the reaction during the given time interval.
[tex]\Delta [x][/tex] = Change in concentration of reactant with respect to time.
[tex]\Delta T[/tex] = Change in time.
[tex]x_1[/tex]=Concentration of reactant at time[tex]t_1[/tex]
[tex]x_2[/tex]=Concentration of reactant at time[tex]t_2[/tex]
So, at [tex]t_1=8.0 sec[/tex] the concentration of [tex]AX_2[/tex] :
[tex]x_1=0.0249 mol/L[/tex]
And at [tex]t_2=2.0 sec[/tex] the concentration of [tex]AX_2[/tex] :
[tex]x_2=0.0088 mol/L[/tex]
The average rate of the reaction at given interval will be given as:
[tex]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[/tex]
0.001341 mol/L s is the average rate from 8.0 to 20.0 seconds.
The average rate from 8.0 to 20.0 seconds in mol / L is 0.001341 mol/L.
Decomposition reactions are those in which a reactant undergoes reaction to form two or more products.
Given,
equation :
[tex]\bold{2 AX_2(g) = 2 AX(g) + X_2(g}[/tex]
t1 = 8.0 s and t2 = 20.0 s
x1 = 0.0249 and x2 = 0.0088
Average rate of reaction:
[tex]\bold{ Ra =-\dfrac{\Delta [x]}{Delta T} = \dfrac{x_2-x_1}{t_2-t_1} }[/tex]
[tex]\bold{ Ra = \dfrac{0.0088-0.0249}{20.0-8.0}= 0.001341 mol\;L }[/tex]
Thus, the average rate is 0.001341 mol/L.
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