[soaring206]

Here is the question I am struggling with:

Equal volumes of two equimolar solutions of reactants A and B are mixed, and the reaction A + B --> C occurs.  At the end of 1 hour, A is 90% reacted.  How much of A will be left unreacted at the end of 2 hours if the reaction is:
a). First order in A and zero order in B
b). First order in A and first order in B
c). Zero order in both A and B

Now, I know that the equation to use for b) is (1/(b[A]₀-a₀))(ln(₀/[A]₀)-ln(_t/[A]_t))=k*t.  However, I wasn't given any concentrations or % concentrations for B, and only that A and B are equimolar.  Also, that equation, as far as I understand, only works when both reactants are first order, so I have no idea where to even begin solving a) and c).  Any help you can give me would be greatly appreciated!  Thank you!

[Donaldson Tan]

all these are plain calculus, using integration by variable seperable technique.

A + B -> C

Ao: initial concentration of A in reaction mixture
Bo: initial concentration of B in reaction mixture

a) First order in A and zero order in B
-d[A]/dt = k[A]
ln [A] = -kt + ln Ao
after 1h, 90% A reacted, left 10% = 0.1
ln 0.1Ao = -k(1) + ln Ao
ln 0.1 + ln Ao = -k + ln Ao
=> k = - ln 0.1
ln [A] = (ln 0.1)t + ln Ao
end of 2h,
ln [A] = (ln 0.1)(2) + ln Ao = ln 0.01 + ln Ao = ln 0.01Ao
1% of A will remain unreacted at end of 2h.

c) Zero order in both A and B
rate of reaction is independent of [A] &
-d[A]/dt = k
[A] = -kt + Ao
end of 1h:
0.1Ao = -k + Ao
=> k= 0.9Ao
[A] = (-0.9Ao)t + Ao
end of 2h:
[A] = (-0.9Ao)(2) + Ao = -0.8Ao
reaction stop when A = 0, ie. all A has reacted.
Zero A present after 2h

[soaring206]

Thank you, that helped a lot!

[Donaldson Tan]

to solve (b), use variable seperable technique for the required calculus to solve the differential equation. express in terms of [A], Ao and Bo.