Question

The reaction AB(aq)→A(g)+B(g) is second order in AB and has a rate constant of 0.0249 L⋅mol−1⋅s−1 at 25.0 ∘C. A reaction vessel initially contains 250.0 mL of 0.180 mol⋅L−1 AB which is allowed to react to form the gaseous product. The product is collected over water at 25.0 ∘C. Part A How much time is required to produce 270.0 mL of the products at a barometric pressure of 736.7 mmHg . (The vapor pressure of water at this temperature is 23.8 mmHg.)

Answer 1

1) we assume both A and B are gases
2) we calculate P(A+B).. (Pressure A + Pressure B)
3) we convert P(A+B) to PA.. (we assume equal concentrations and kinetic energy so that PA = PB)
4) we convert PA to moles A
5) we convert moles A to moles AB
6) we calculate how much AB must have reacted
7) use the integrated 2nd order rate equation to solve for t

here we go...
P(A+B) = Ptotal - PH2O = 736.7mmHg - 23.8mmHg = 712.9mmHg
PA = 1/2 P(A+B) = 356.45mmHg
nA = PA x V/(RT) = 356.45mmHg x (1atm/760mmHg) x 0.27L / (0.08206 Latm/moleK x 298.15K)
nA = 0.005175 moles A

then..
from the balanced equation, 1 mole AB ---> 1 mole A
so. moles AB reacted = 0.005175 moles

and...
initial moles AB = 0.2500L x (0.180 moles / L) = 0.045
remaining moles AB = 0.045 - 0.005175 = 0.03982
final molarity = 0.03982moles / 0.2500L = 0.1593M

and from here...assuming second order...
- d[A] / dt = k x [A]²
1 / [A]² d[A] = -k dt
∫1 / [A]² d[A] = ∫-k dt...
-1/[At] + 1/[Ao] = -kt
1/[At] = +kt + 1/[Ao]

so..
t = (1/[At] - 1/[Ao]) / k
Time t = (1/0.1593M - 1/0.180M) / (0.0249 / Msec) = 28.99sec