The rate constant for the first order decomposition of gaseous N2O5 to NO2 and O

Question

The rate constant for the first order decomposition of gaseous

N2O5 to NO2 and O2: N2O5(g) 2NO2(g) + 1/2O2(g)

is 1.7x103 s 1 at 55°C.

a) If 2.70 g of gaseous N2O5 is introduced to an evacuated 2.00 L container maintained at a constant temperature of 55°C, what is the total pressure in the container after reaction time of 13.0 min?

b) Use the data in Appendix B to calculate the initial rate at which the reaction mixture absorbs heat (in J/s). You may assume that the heat of the reaction is independent of temperature.

(N2O5: heat of formation = 13.3 kJ/mol; G of formation = 117.1 kJ/mol; S = 355.6 J/K*mol)

c) What is the total amount of heat absorbed (in kJ) after a reaction time of 10.0 min?

Explanation / Answer

N2O5(g)        2NO2(g) + 1/2 O2(g)

K = 1.7x10-3 s-1

(a). Mass of N2O5 = 2.70 g

Molar mass of N2O5 = 108.01 g/mol

Moles of N2O5 = 2.70 / 108.01

= 0.025

Volume of container = 2.0 L

Temperature = 55 oC = 328 K

Using PV = nRT

P * 2.0 = 0.025 * 0.0821 * 328

P = 0.336 atm

So initial pressure, Po = 0.336 atm

Time, t = 13.0 min = 780 sec

Using first order rate law:

ln P = - kt + ln Po

ln P = - 1.7x10-3 * 780 + ln 0.336

ln P = - 1.326 - 1.09

ln P = - 2.416

P = 0.089 atm

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