Question

Sulfur dioxide, SO2(g), can react with oxygen to produce sulfur trioxide, SO3(g), by the following reaction
2SO2+O2=2SO3

The standard enthalpies of formation for SO2(g) and SO3(g) are
deltaH= SO2(g)= -296.8 kj
dH= SO3(g)= -395.7 kJ
Calculate the amount of energy in the form of heat that is produced when a volume of 3.75 L of SO2(g) is converted to 3.75 L of SO3(g) according to this process at a constant pressure and temperature of 1.00 atm and 25.0

Answer 1

Answer:

The heat produced is -15,1kJ

Explanation:

For the reaction:

2SO₂+O₂ → 2SO₃

The enthalpy of reaction is:

ΔHr = 2ΔHf SO₃ - 2ΔHf SO₂

As ΔHf SO₃ = -395,7kJ and ΔHf SO₂ = -296,8kJ

ΔHr = -197,8kJ

Using n=PV/RT, the moles of reaction are:

$n = \frac{1,00atm*3,75L}{0,082atmL/molK*298,15K}$ = 0,153 moles of reaction

As 2 moles of reaction produce -197,8kJ of heat, 0,153moles produce:

0,153mol×$\frac{-197,8kJ}{2mol}$ = -15,1kJ

I hope it helps!

Answer 2

14.8 KJ/mol of heat energy is produced when a volume of 3.75 L of SO2(g) is converted to 3.75 L of SO3(g).

The equation of the reaction is;

2SO2(g) + O2(g) ⇄ 2SO3(g)

We have the following information;

ΔHf SO2(g) = -296.8 KJ/mol

ΔHf O2(g) = 0 KJ/mol

ΔHf SO3(g) =  -395.7 kJ/mol

We can calculate the heat of reaction ΔHrxn from;

ΔHrxn = [ΔHf (products) - ΔHf (reactants)]

ΔHrxn = [2( -395.7 kJ/mol)] - [2(-296.8 KJ/mol) + 0 KJ/mol]

ΔHrxn = (-791.4  kJ/mol) + 593.6 KJ/mol

ΔHrxn = -197.8 KJ/mol

We can find the number of moles of SO2 reacted using the ideal gas equation;

P = 1.00 atm

T = 25.0°C + 273 = 298 K

n = ?

V = 3.75 L

R = 0.082 atmLK-1mol-1

So,

PV = nRT

n = PV/RT

n = 1.00 atm × 3.75 L/0.082 atmLK-1mol-1 ×   298 K

n = 0.15 moles

If 2 moles of SO2 produced -197.8 KJ/mol

0.15 moles of SO2 will produce  0.15 moles  × (-197.8 KJ/mol)/ 2 moles

= -14.8 KJ/mol

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