Gaseous iodine (I2 (g)) dissociates into iodine atoms at 625 K with a rate constant of 0.271 s-1.

Question

If you start with 0.0560 M I₂, what is the initial pressure of I₂ at 625 K?

Answer 1

Not quite sure what the rate constant has to do with the initial pressure. Usually with a gas, you can find the pressure from the ideal gas law, PV = nRT but we don't know the volume. If the question were to ask for the concentration of I2 at some time point, we could use the rate constant to find t_1/2 and then use the integrated rate law to find the [ ] (see below)

For a 1st order reaction the half life (t_1/2) = ln 2 / k

Therefore, t_1/2 = 0.693 / 0.271 s^-1 = 2.56 s

Integrated rate law: ln [A] = -kt + ln [A]_o

If we assume 1 L volume, then we can find the initial pressure of I₂(g), but usually the volume is given.

PV = nRT

P = nRT / V

P = pressure = ?

n = moles = 0.0560

R = gas constant = 0.0821 Latm/Kmol

V = volume = 1 L (???)

T = temperature in Kelvin = 625K

P = (0.0560)(0.0821)(625) / 1

P = 2.87 atm