At 65 degree C, the half-life for the first-order decomposition of N_2O_5 is 2.3

Question

At 65 degree C, the half-life for the first-order decomposition of N_2O_5 is 2.38 min. N_2O_5 (g) rightarrow 2NO_2 (g) + 1/2 O_2 (g) 1.50 g of N_2O_5 is introduced into an evacuated 16 - L flask at 65 degree C. What is the initial partial pressure, in mmHg, of N_2O_5 (g)? Express your answer using two significant figures. What is the partial pressure, in mmHg, of N_2O_5 (g) after 2.38 min? Express your answer using two significant figures. What is the total gas pressure in mmHg, after 2.38 min? Express your answer using two significant figures.

Explanation / Answer

A)

Partial Pressure of N2O5 is given by

P-N2O5 = nRT/V

n = moles = mass/MW = 1.5/108.01 = 0.0138876 moles of N2O5

P-N"O5 = 0.0138876*62.3*(65+273)/(16) = 18.277 mm Hg of N2O5

B)

after 2.38 mins, find P-N2O5

recall that this is HALF life

so

P must be reduced at 1/2 of its original value

so

P-N2O5 = 1/2*18.277 = 9.1385 mm Hg

then

Find Total P

N2O4 = x = 9.1385

NO2 = 2*x = 2*9.1385 = 18.277

O2 = 1/2*x =1/2*9.1385 = 4.56925

Total P = 9.1385+18.277+4.56925 = 31.98475 mm Hg

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