Question

The reaction AB(aq)→A(g)+B(g) is second order in AB and has a rate constant of 0.0122 M−1⋅s−1 at 25.0 ∘C. A reaction vessel initially contains 250.0 mL of 0.180 M AB which is allowed to react to form the gaseous product. The product is collected over water at 25.0 ∘C How much time is required to produce 134.0 mL of the products at a barometric pressure of 763.7 mmHg . (The vapor pressure of water at this temperature is 23.8 mmHg.)

Answer 1

Initially A 250 ml of 0.180M

number of moles of AB initially = 0.180M x 0.250L=0.045 moles

T= 25C = 25+273= 298K

V= 134ml= 0.134L

total pressure = 763.7mm

vapour pressure of water= 23.8mm

pressure of dry gas= 763.7 - 23.8 = 739.9mm

P= 739.9/760 =0.974 atm

R= 0.0821 L-atm/mol-K

Ideal gas equation

PV= nRT

n= PV/RT = 0.974 x 0.134/0.0821x298 =0.0053 moles

final number of AB= 0.0053 moles

for second order reaction

1/[A] = Kt + 1/[A]0

1/[A] - 1/[A]0 = Kt

1/0.0053 - 1/0.045 = 0.0122 xt

188.68 -22.22 = 0.0122 xt

t= 13644.26 sec.