Answer:
The rate is 0.038 M/s
Explanation:
The reaction follows a zero order
Let the concentration of A after 6.5 minutes be y
Rate = k[y]^0 = change in concentration/time
k (rate constant) = 3.8×10^-2 mol/L.s
Initial concentration of A = 1×10^-3 M
Change in concentration = 1×10^-3 - y
Time = 6.5 minutes = 6.5×60 = 390 seconds
3.8×10^-2×1 = 1×10^-3 - y/390
0.001 - y = 0.038×390
0.001 - y = 14.82
y = 0.001 - 14.82 = -14.819 M
Rate = change in concentration/time = [0.001 - (-14.819)]/390 = (0.001+14.819)/390 = 14.82/390 = 0.038 M/s
The rate after 6.5 minutes is 0.038m/s.
This is a zero order reaction.
We can depict the concentration of A after 6.5 minutes be x
Rate = k[y]⁰ = change in concentration/time
Rate constant is represented as k = 3.8×10⁻² mol L⁻¹ s⁻¹
Initial concentration of A = 1×10⁻³ M
Change in concentration = 1×10⁻³ - x
The rate after 6.5 minutes
Time = 6.5 minutes = 6.5×60 = 390s.
3.8×10⁻² ×1 = 1×10⁻³ - x/390
0.001 - x = 0.038×390
0.001 -x = 14.82
x = 0.001 - 14.82 = -14.819 M
Rate = change in concentration/time
= [0.001 - (-14.819)]/390 = (0.001+14.819)/390
= 14.82/390
= 0.038 M/s.
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