Question

Sulfur dioxide and oxygen react exothermically according to the thermochemical equation shown below. Calculate the maximum amount of heat that could be released if 15.0 g of O2 gas is allowed to react with 13.4 L of SO2 gas at 1.00 atm and 15°C. (CAUTION: This problem may not be as simple as it looks. So please think about it before you start working.) 2 SO2(g) + O2(g) → 2 SO3(g) ΔH° = –198 kJ

Answer 1

According to reaction 2 mole of SO₂ react with 1 mole of O₂ to release heat -198 KJ

first callculate mole of SO₂ by using ideal gas equation

Ideal gas equation

PV = nRT             where, P = 1 atm,

V = volume in Liter = 13.4 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 15⁰C = 273.15+ 15 = 288.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (1 13.4)/(0.08205 288.15) = 0.56677 mole of SO₂

molar mass of O₂ = 31.9988 gm/mol that mean 1 mole of O₂ = 31.9988 gm

then 15.0 gm of O₂ = 15/31.9988 = 0.468767 mole

According to reaction 2 mole of SO₂ react with 1 mole of O₂ then to react with 0.468767 mole of O₂ require

0.468767 2 = 0.967535 mole of SO₂ but SO₂ given only 0.56677 mole therefore SO₂ is limiting reactant react completly

2 mole of SO₂ react release heat -198 KJ then 0.56677 mole of SO₂ release heat =

0.56677 (-198) /2 = -56.11 KJ

maximum amount of heat released = -56.11 KJ