Hydrogen iodide decomposes slowly to H2 and I2 at 600K . The reaction is second order in HI and the rate constant is 9.7 x 10^-6 M^-1s^-1 . If the initial concentration of HI is 0.110 M.
What is its molarity after a reaction time of 7.00 days?
What is its molarity after a reaction time of 7.00 days?
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rate = -d[A] / dt = k x [A]^n
if n = 2.. ie 2nd order rxn in A.. then..
-d[A] / dt = k x [A]²
if we have 2 (t, [A]) points.. (0, [Ao]) and (t, [At])... ie we start with concentration [Ao] at time = 0 and it drops to [At] after time = t has elapsed..we can rearrange and integrate to get
1/[A]² d[A] = -k dt
∫1/[A]² d[A] = -k ∫dt
-1/[At] - -1/[Ao] = -k x (t - 0)
1/[At] = kt - 1/[Ao]
solving..
1/[At] = (9.7x10^-6 /M/s) x (7.00days x 24hr/day x 3600s/hr) - 1/(0.110M)
[At] = ?....you get to finish
if n = 2.. ie 2nd order rxn in A.. then..
-d[A] / dt = k x [A]²
if we have 2 (t, [A]) points.. (0, [Ao]) and (t, [At])... ie we start with concentration [Ao] at time = 0 and it drops to [At] after time = t has elapsed..we can rearrange and integrate to get
1/[A]² d[A] = -k dt
∫1/[A]² d[A] = -k ∫dt
-1/[At] - -1/[Ao] = -k x (t - 0)
1/[At] = kt - 1/[Ao]
solving..
1/[At] = (9.7x10^-6 /M/s) x (7.00days x 24hr/day x 3600s/hr) - 1/(0.110M)
[At] = ?....you get to finish