The reaction 2A → A2 was experimentally determined to be second order with a rate constant, k, equal to 0.0265 M–1min–1. If the initial concentration of A was 3.00 M, what was the concentration of A (in M) after 180.0 min?
M ?
Given, the balanced second order reaction,
2A A2
Also given,
Rate constant(k) = 0.0265 M-1 min-1
The initial concentration of A (No) = 3.00 M
Time (t) = 180.0 min
Nt = ? The concentration of A(in M) after time "t".
We know, The second order rate law equation,
1 / [Nt] = kt + 1 / [No]
Substituting the known values in the formula,
1 / [Nt] = 0.0265 M-1 min-1 x 180.0 min. + 1 / 3.00 M
1 / [Nt] = 4.77 + 0.3333
1 / [Nt] = 5.1033
[Nt] = 1 / 5.1033
[Nt] = 0.196 M
The reaction 2A → A2 was experimentally determined to be second order with a rate constant,...
The reaction 2A → A2 was experimentally determined to be second order with a rate constant, k, equal to 0.0265 M–1min–1. If the initial concentration of A was 5.75 M, what was the concentration of A (in M) after 180.0 min?
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