Question

The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10-3min-1. If a rigid vessel initially contains only N2O5 at a pressure of 125 kPa, how long will it take for the total pressure to reach 176 kPa?
a. 113 min
b. 129 min
c. 42 min
d. 182 min
e. 62 min
f. 83 min

Answer 1

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

                   2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0        125 kPa

At t = teq     125 - 2x      4x        x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $2.8\times 10^{-3}$ min⁻¹

Initial concentration $[A_0]$ = 125 kPa

Final concentration $[A_t]$ = 91 kPa

Time = ?

Applying in the above equation, we get that:-

$91=125e^{-2.8\times 10^{-3}\times t}$

$125e^{-2.8\times \:10^{-3}t}=91$

$-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)$

$t=113\ min$