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.

Respuesta :

Answer:

  • 39.0 min

Explanation:

Second order decomposition (1 reactant) reaction means:

  • rate = - d {HI] / dt = Ā k [HI]Ā² =

  • Integrating:

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

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

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

[tex]\frac{1}{P_t}-\frac{1}{P_0}=kt[/tex]

Here you have:

  • t = ?
  • P at t = 0 = 0.691 atm
  • P at t = t = 0.187 atm
  • K = 0.100 atmā»Ā¹ . minā»Ā¹

Therefore, substituting:

  • 1 / 0.187atm - 1 / 0.691atm = Ā 0.100 atmā»Ā¹ . minā»Ā¹ . t

  • t = 3.900atmā»Ā¹ / Ā 0.100 atmā»Ā¹ minā»Ā¹

  • t = 39.0 min ā† answer