How long will it take for the concentration of NO2 to decrease?

Question

The decomposition of NO2(g) occurs by the following bimolecular elementary reaction.

2NO2(g) → 2 NO(g) + O2(g)

The rate constant at 273 K is 2.3 10-12 L/mol · s, and the activation energy is 111 kJ/mol. How long will it take for the concentration of NO2(g) to decrease from an initial partial pressure of 5.0 atm to 1.8 atm at 471 K? Assume ideal gas behavior.

Answer 1

Hi Abby! This is a loaded question, so lets break this down by parts.

Since this is a second-order reaction and we need to determine time from concentration, we must use the integrated rate law for a second-order reaction:

1/[A]t - 1/[A]0 = kt

Remember the rate constant is dependent of temperature. Since they are asking for the reaction at 471K, let us first find the rate constant at 471K using the Arrhenius equation:

k = Ae^(-Ea/RT), where A is the pre-exponential factor, Ea is the activation energy, R is the thermodynamic rate constant (0.008314 kJ/mol-K) and T is the temperature.

We know that k(273) = 2.3 x 10^-12. We also know the activation energy (which changes slightly with temperature, but is negligible). Therefore, we can set up two equations:

2.3 x 10^-12 = Ae^(-111 kJ/mol / (0.008314 kJ/mol K * 273K)

k(471) = Ae^(-111kJ/mol / (0.008314 kJ/mol K * 471 K)

If we take the two equations and divide them, we can cancel out A, and notice then that our only unknown is k(471). I will now simplify the exponential terms:

2.3 x 10^-12/k(471) = 5.767 x 10^-22 / 4.89 x 10^-13

Simplifying the right side gives:

2.3 x 10^-12/k(471) = 1.179 x 10^-9

And solving for k(471) gives:

2.3 x 10^-12/1.179 x 10^-9 = k(471) = 1.9 x 10^-3 L/mol-s

Now, the next part is to determine the concentration value in M, since our rate constant is in 1/M-s. Since we are assuming and ideal gas, we can determine concentration from the ideal gas equation:

PV=nRT --> n/V = P/RT --> M= P/RT

Therefore, [A]0 = 5.0/ (0.08206*471) = 0.129 M and [A]t = 1.8/(0.08206 * 471) = 0.0466 M

Finally, we can use the rate constant at 471 K and the concentration values to determine how long the reaction will take:

1/0.0466M - 1/0.129 M = (1.9 x 10^-3 1/M-s) t

13.707/M = 1.9 x 10^-3/M-s * t

t = 13.707/(1.9 x 10^-3) s

t = 7214 s

I hope this helps!