 # Enthalpy Change Calculations for a Chemical Reaction Tutorial

## Key Concepts

• Enthalpy is also known as heat content and has the symbol H.
• The enthalpy change, or change in heat content, for a chemical reaction refers to the heat that this absorbed or released during the reaction.
• The symbol given to a macroscopic change is Δ (capital Greek letter delta)
• The symbol for the enthalpy change (change in heat content) for a chemical reaction is therefore ΔH
• Most commonly, the units given for enthalpy change, ΔH, are kilojoules per mole, kJ mol-1

ΔH = ? kJ mol-1

• The value of the enthalpy change can be + (positive) or - (negative)

endothermic reaction (heat absorbed) : ΔH is positive, ΔH = + ? kJ mol-1

exothermic reaction (heat released) : ΔH is negative, ΔH = - ? kJ mol-1

• The value of ΔH for a given reaction can be:

(a) determined experimentally using a cup calorimeter or an electrical calorimeter

(b) calculated from bond energies or heats of formation

• If a chemical equation is reversed, the sign of the enthalpy change value, ΔH, is also reversed.

⚛ For the forward reaction: X + 2Y → XY2     ΔH = -h kJ mol-1

⚛ For the reverse reaction: XY2 → X + 2Y     ΔH = +h kJ mol-1

• The value of ΔH given in kJ mol-1 refers to kJ per 1 mole of reactant or product as written in the balanced chemical equation.

For the reaction: X + 2Y → XY2     ΔH = - h kJ mol-1

h kJ of energy released for every 1 mole of X consumed.

h kJ of energy released for every 2 moles of Y consumed.

h kJ of energy released for every 1 mole of XY2 produced.

• The energy released or absorbed during a chemical reaction (energy) can be calculated using the stoichiometric coefficients (mole ratio) from the balanced chemical equation and the value of the enthalpy change for the reaction (ΔH):

 energy = n × ΔH           stoichiometric coefficient

where

energy = energy released or absorbed measured in kJ
n = amount of substance measured in moles
ΔH = enthalpy change for the reaction measured in kJ mol-1
stoichiometric coefficient is that for the particular reactant or product of interest

 For the reaction: stoichiometriccoefficient(mole ratio) energy releasedper mole of reagent energy releasedper n molesof reagent X + 2Y → XY2 ΔH = -h kJ mol-1 1 2 1 h/1 = h h/2 = ½ × h h/1 = h n × h n × h/2 n × h

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## Enthalpy Change When Chemical Equations Are Reversed

If a chemical equation is reversed, the sign of the ΔH value is also reversed.

Consider the chemical reaction for the synthesis of ammonia gas (NH3(g)) from nitrogen gas (N2(g)) and hydrogen gas (H2(g)) as shown in the balanced chemical equation below:

NH3(g) synthesis: N2(g) + 3H2(g) → 2NH3(g)     ΔH = -92.4 kJ mol-1

The reaction is exothermic, energy is released during the reaction, so the enthalpy change term (ΔH) is negative (-).
Because energy is released, we can think of energy as being a product of the chemical reaction.
So we could write this equation as:

NH3(g) synthesis: N2(g)+ 3H2(g) → 2NH3(g) + 92.4 kJ mol-1

But what if we wanted to decompose ammonia gas into nitrogen gas and hydrogen gas?
The decomposition of ammonia is the reverse of the synthesis of ammonia gas shown above.
So, we can write the synthesis reaction equation in reverse to give the decomposition reaction:

NH3(g) decomposition: 2NH3(g) + 92.4 kJ mol-1 → N2(g) + 3H2(g)

For the decomposition reaction, energy is a reactant, energy is absorbed during the reaction.
The decomposition reaction is endothermic so the sign of the enthalpy change term (ΔH) will be positive (+).
We can write the balanced chemical equation for the decomposition reaction as:

NH3(g) decomposition: 2NH3(g) → N2(g) + 3H2(g)     ΔH = + 92.4 kJ mol-1

If we compare the two reactions, synthesis and decomposition, side-by-side, we see that when we reverse the chemical equation we also reverse the sign of the ΔH value for the reaction:

 synthesis of ammonia: decomposition of ammonia: N2(g) + 3H2(g) → 2NH3(g) ΔH = -92.4 kJ mol-1 2NH3(g) → N2(g) + 3H2(g) ΔH = +92.4 kJ mol-1

We can generalise and say that the sign of ΔH for the forward reaction is the opposite sign of ΔH for the reverse reaction.
For any reaction:

Forward reaction: X → Z     ΔH = +h kJ mol-1

Reverse reaction: Z → X     ΔH = -h kJ mol-1

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## Calculating Enthalpy Change For a Specific Amount of Reactant or Product

The value of ΔH given as kJ mol-1 refers to kJ per 1 mole of reactant or product as written in the equation.

For example, the synthesis of ammonia gas (NH3(g)) from nitrogen gas (N2(g)) and hydrogen gas (H2(g)) releases 92.4 kJ mol-1 of heat energy as shown by the balanced chemical equation below:

N2(g) + 3H2(g)2NH3(g)     ΔH = - 92.4 kJ mol-1

This means that:

• 92.4 kJ of energy is released for every 1 mole of N2(g) consumed.
• 92.4 kJ of energy is released for every 3 moles of H2(g) consumed.
• 92.4 kJ of energy is released for every 2 moles of NH3(g) produced.

But what if we start this reaction with 10 moles of N2(g) and 30 moles of H2(g), how much energy would be released then?

If 1 mole of N2(g) produced 92.4 kJ of energy, then 10 times as much N2(g) (10 moles of N2(g)) should produce 10 times as much energy:

10N2(g) + 30H2(g)20NH3(g)   energy = 10 × ΔH

energy released = 10 × ΔH
= 10 mol × 92.4 kJ mol-1
= 924 kJ

But what if we use 10 moles of H2(g) and an excess of N2(g) ? How much energy will be released when ammonia gas is produced then?

It is useful to consider first how much energy would be released by just 1 mole of H2(g).
We can do this by dividing the entire chemical equation by the stoichiometric coefficient of H2(g) in the chemical equation, that is, divide every term in the chemical equation (including the value of ΔH!) by 3:

 original chemical equation divide by 3 N2(g) + 3H2(g) → 2NH3(g) ΔH = -92.4 kJ mol-1 1/3N2(g) + 3/3H2(g) → 2/3NH3(g) ΔH = -92.4/3 kJ mol-1 1/3N2(g) + H2(g) → 2/3NH3(g) ΔH = -30.8 kJ mol-1

From this new balanced chemical equation we see that when 1 mole of H2(g) is consumed (with excess N2(g)) to produce 2/3 moles of NH3(g), 30.8 kJ of energy will be released.

If 1 mole H2(g) produced 30.8 kJ of energy, then 10 times the amount of H2(g) will produce 10 times as much energy:

energy released = 10 × 30.8 = 308 kJ

If we think about what we have done here we have:
(a) divided ΔH value by the stoichiometric coefficient of H2(g) in the balanced chemical equation
(b) multiplied this new value by the moles of H2(g) actually present

We can generalise this for any balanced chemical equation:

aA + bBcC + dD     ΔH = +h kJ mol-1

To calculate the energy absorbed for n moles of reactant A:

(a) divide ΔH value by a
energy absorbed per mole of A = ΔH/a     kJ mol-1

(b) then multiply by n :
energy absorbed for n moles of A = n × ΔH/a     kJ

To calculate the energy absorbed for n moles of reactant B:

(a) divide ΔH value by b
energy absorbed per mole of B = ΔH/b     kJ mol-1

(b) then multiply by n :
energy absorbed for n moles of B = n × ΔH/b     kJ

To calculate the energy absorbed for n moles of product C:

(a) divide ΔH value by c
energy absorbed per mole of C = ΔH/c     kJ mol-1

(b) then multiply by n :
energy absorbed for n moles of C = n × ΔH/c     kJ

To calculate the energy absorbed for n moles of product D:

(a) divide ΔH value by d
energy absorbed per mole of D = ΔH/d     kJ mol-1

(b) then multiply by n :
energy absorbed for n moles of D = n × ΔH/d     kJ

The energy released or absorbed during a chemical reaction (energy) can be calculated using the stoichiometric coefficients (mole ratio) from the balanced chemical equation and the value of the enthalpy change for the reaction (ΔH):

 energy = n × ΔH           stoichiometric coefficient

where

energy = energy released or absorbed measured in kJ
n = amount of substance measured in moles
ΔH = enthalpy change for the reaction measured in kJ mol-1
stoichiometric coefficient is that for the particular reactant or product of interest

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## Worked Example: Calculating Heat Released or Absorbed for a Given Amount of Reactant or Product

Question: When sulfur trioxide gas (SO3(g)) is synthesised from sulfur dioxide gas (SO2(g)) and oxygen gas (O2(g)), 198 kJ mol-1 of heat is released as shown in the balanced chemical equation below:
2SO2(g) + O2(g) → 2SO3(g)     ΔH = -198 kJ mol-1
Determine the amount of heat absorbed in kilojoules when 20.0 g moles of sulfur trioxide gas decomposes to produce sulfur dioxide gas and oxygen gas.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate the heat absorbed in kJ for the decomposition of 20.0 g SO3(g)
heat absorbed = ? kJ

2. What data (information) have you been given in the question?

Extract the data from the question:

Balanced chemical equation for the synthesis of SO3(g):
2SO2(g) + O2(g) → 2SO3(g)     ΔH = -198 kJ mol-1

m(SO3(g)) = mass of sulfur trioxide gas = 20.0 g

3. What is the relationship between what you know and what you need to find out?

(i) Reaction for decomposition of SO3(g) is the reverse of the reaction for synthesis of SO3(g) so the sign of the ΔH term is reversed:
ΔHdecomposition = -ΔHsynthesis

(ii) (a) Calculate amount in moles of SO3(g):
n(SO3(g)) = moles of SO3(g) = mass(SO3(g)) ÷ molar mass (SO3(g))

(b) Calculate energy absorbed by decomposition of n moles of SO3(g):

 energy absorbed = n(SO3(g)) × ΔH              stoichiometric coefficient of SO3(g)

4. Perform the calculations
(i) ΔHdecomposition = -ΔHsynthesis = -(-198) kJ mol-1 = +198 kJ mol-1
Decomposition reaction: 2SO3(g) → 2SO2(g) + O2(g)     ΔH = + 198 kJ mol-1

(ii) (a) Calculate moles of SO3(g)
n(SO3(g)) = mass(SO3(g)) ÷ molar mass(SO3(g))
mass(SO3(g)) = 20.0 g
molar mass = 32.06 + 3 × 16.00 = 80.06 g mol-1 (from periodic table)
n(SO3(g)) = 20.0 g ÷ 80.06 g mol-1 = 0.2498 mol

(b) Energy absorbed by decomposition of 0.2498 moles of SO3(g):
n = 0.2498 mol
stoichiometric coefficient of SO3(g) from balanced chemical equation is 2

 energy absorbed = 0.2498 mol × 198 kJ mol-1 2 = 24.73 kJ = 24.7 kJ

Work through the problem, step-by-step:

198 kJ mol-1 of energy is released during the synthesis of SO3(g), that is, heat is a product of the synthesis reaction:

2SO2(g) + O2(g) → 198 kJ mol-1 + 2SO3(g)

The decomposition of SO3(g) is the reverse of this reaction:

198 kJ mol-1 + 2SO3(g) → 2SO2(g) + O2(g)

Decomposition of 2 moles of SO3(g) absorbs 198 kJ of heat.
Divide every term in the equation by 2 so that we have only 1 mole of SO3(g) in the equation:

198/2 kJ mol-1 + 2/2SO3(g)2/2SO2(g) + 1/2O2(g)

99 kJ mol-1 + SO3(g) → SO2(g) + ½O2(g)

Therefore decomposition of 1 mole of SO3(g) absorbs 99 kJ of heat

How many moles of SO3(g) do we have?
We have 20 g of SO3(g):
moles of SO3(g) = 20 g ÷ (32.06 + 3 × 16) = 0.2498 mol
This is about 1 quarter of a mole, so we expect the energy that will be absorbed will be about 1 quarter of the energy absorbed for 1 mole:
energy absorbed ≈ ¼ × 99 kJ = 24.75 kJ

Since this answer is about the same as the one we calculated above, we are reasonably confident that our answer is plausible.

6. State your solution to the problem "heat absorbed in kJ to decompose 20.0 g SO3(g)":

heat absorbed = 24.7 kJ

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