The rate constant for the low pressure decomposition of N₂O₄(g) is 3.60 × 10⁴ L･mol⁻¹･s⁻¹ at 250 K. If the

Question

The rate constant for the low pressure decomposition of N₂O₄(g) is 3.60 × 10⁴ L･mol⁻¹･s⁻¹ at 250 K. If the initial pressure of N₂O₄(g) is 35.0 torr at 250 K, what is the pressure of N₂O₄(g) remaining after 15.0 milliseconds? Assume any reverse reactions are insignificant under these conditions. Note: 760 torr/1 atm.

Answer 1

Hi Terra H.!

Step 1: Determine Order of Reaction

We can determine the order of reaction from the rate constant given.

The equation for the nth order unit of the rate constant is as follows:

k = L^n-1*mol^1-n*s^-1

Let's compare our given rate constant with our equation above. The question tells us the rate constant is:

k = 3.60x10⁴ L*mol^-1*s^-1

This tells us the following information:

1. Units of volume is to the first power
2. Units of moles is to the negative first power

If we use our general equation above, we can determine the order of the reaction.

Volume is given as L^n-1

Our rate constant has volume raised to the first power.

So:

n - 1 = 1 => n = 2

We can use moles as a way to double-check our conclusion above.

Moles are given as mol^1-n

Our rate constant has moles raised to the negative first power.

So:

1 - n = -1 => n = 2

We can deduce that this is a decomposition reaction that follows 2nd order kinetics.

Step 2: Find the Initial Concentration

We are given that the initial pressure of N₂O₄ is equal to 25 torr:

P_o = 25 torr

Let's use the ideal gas law to determine the initial concentration when the pressure is equal to P₀.

PV = nRT => P = (n/V)RT

=> n/V = C

P = CRT => C = P/RT

We know that the units of the gas constant, R, are equal to L*atm*mol^-1*K^-1

So, we must convert from torr into atm

(25 torr)(1 atm/760 torr)

P_o = 0.033 atm

C = P/RT

<=> 0.033atm/[(0.0821 L*atm*mol^-1*K^-1)(250K)]

C_o = 0.0016 mol*L^-1

Step 3: Find the Concentration After t = 15ms

Now that we have determined the initial concentration, we must ascertain the final concentration after t = 15 ms.

Let's use the 2nd order integrated rate law to make this determination.

1/C = 1/C_o + kt

<=> (1/0.0016 mol*L^-1) + (3.60x10⁴ L*mol^-1*s^-1)(15x10^-3s)

<=> 623.96 L*mol^-1 + 540 L*mol^-1

1/C = 1163.96 L*mol^-1

C = 8.59x10^-4 mol*L^-1

Note that we first obtain the inverse of our answer, which is why it is recorded in L*mol^-1.

Step 4: Find the Final Pressure

Once we have determined the final concentration, we can then determine what final pressure it exerts. Let us once again turn to the ideal gas law.

P_f = C_fRT

<=> (8.59x10^-4 mol*L^-1)(0.0821 L*atm*mol^-1*K^-1)(250K)

P_f = 0.0176 atm = 13.4 torr

Hope this helps!

Cheers