Question

HI(g) decomposes to molecular hydrogen and molecular iodine by a
second-order reaction. If the initial pressure of Hl is 0.691 atm and
the rate constant for the reaction is 0.100 atm-1.min-1, how much
time (min) will pass before the HI pressure drops to 0.187 atm?
Enter your answer to 1 decimal place.

Answer 1

Answer:

• 39.0 min

Explanation:

Second order decomposition (1 reactant) reaction means:

• rate = - d {HI] / dt =  k [HI]² =

• Integrating:

$\int\limits^{[HI_1]}_{[HI_0]} {\,\frac{d[HI]}{[HI]^2}=-\int\limits^t_0 {t} \, dt$

$\frac{1}{[HI_t]}-\frac{1}{[HI_0]}=kt$

When temperature and volume are kept constant concentration is proportional to pressure (P) and the rate may be expressed in terms of pressure as:

$\frac{1}{P_t}-\frac{1}{P_0}=kt$

Here you have:

• t = ?

• P at t = 0 = 0.691 atm

• P at t = t = 0.187 atm

• K = 0.100 atm⁻¹ . min⁻¹

Therefore, substituting:

• 1 / 0.187atm - 1 / 0.691atm =  0.100 atm⁻¹ . min⁻¹ . t

• t = 3.900atm⁻¹ /  0.100 atm⁻¹ min⁻¹

• t = 39.0 min ← answer