number of molecules of SO2Cl2 remaining after 12.5 h

Question

Sulfuryl chloride undergoes first- order decomposition at 320°C with a half-life of 8.75 h

SO₂Cl₂ (g) à SO₂ (g) + Cl₂ (g)

a.) What is the value of the rate constant, k, in 1/sec?

b.) If the initial pressure of SO₂Cl₂ is 791 torr and the decomposition occurs in a 1.25 L container, how many molecules of SO₂Cl₂ remain after 12.5 h?

Answer 1

First order decay/decomposition demonstrates the following relationships:

t1/2 = 0.692/k

fraction remaining = 0.5n where n = number of half lives that have elapsed

a) t1/2 = 0.693/k

8.75 hr = 0.693/k

8.75 hr x 60 min/hr x 60 sec/min = 0.693/k

31,500 sec = 0.693/k

k = 2.2x10^-5 s^-1

b) Initial moles SO2Cl2: use PV = nRT and solve for n

n = PV/RT = (791 torr)(1.25 L) / (62.36 Ltorr/Kmol)(298K) ASSUMING A TEMPERATURE OF 298K/25ºC

n = 0.0532 moles SO2Cl2 initially present

fraction remaining = 0.5ⁿ where n = number of half lives elapsed = 12.5 hr/8.75 hr = 1.43 half lives elapsed

fraction remaining = 0.5^1.43

fraction remaining = 0.371

moles remaining = 0.371 x 0.0532 moles = 0.0197 moles remaining

molecules remaining = 0.0197 moles x 6.02x10²³ molecules/mole = 1.19x10²² molecules remaining