Introduction to Enthalpy Change Calculations for Chemical Reactions Tutorial

Key Concepts

• The value of the enthalpy change for a chemical reaction, ΔH, can be + (positive) or − (negative) ⁽¹⁾

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

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

• 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 kilojoules of energy per 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 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:	X	+	2Y	→	XY2	ΔH = −h kJ mol-1
  stoichiometric coefficient (mole ratio)	1		2		1
  energy released per mole of reagent	h/1 = h		h/2 = ½ × h		h/1 = h
  energy released per n moles of reagent	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 (NH_3(g)) from nitrogen gas (N_2(g)) and hydrogen gas (H_2(g)) as shown in the balanced chemical equation below:

NH_3(g) synthesis: N_2(g) + 3H_2(g) → 2NH_3(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 with energy shown as a product of the reaction:

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:

NH_3(g) decomposition: 2NH_3(g) + 92.4 kJ mol^-1 → N_2(g) + 3H_2(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:

NH_3(g) decomposition: 2NH_3(g) → N_2(g) + 3H_2(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:

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

Calculating Enthalpy Change For a Specific Amount of Reactant or Product

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

For example, the synthesis of ammonia gas (NH_3(g)) from nitrogen gas (N_2(g)) and hydrogen gas (H_2(g)) releases 92.4 kJ mol^-1 of heat energy as shown by the balanced chemical equation below:

N_2(g) + 3H_2(g) → 2NH_3(g)     ΔH = −92.4 kJ mol^-1

This means that:

• 92.4 kJ of energy is released for every 1 mole of N_2(g) consumed.
• 92.4 kJ of energy is released for every 3 moles of H_2(g) consumed.
• 92.4 kJ of energy is released for every 2 moles of NH_3(g) produced.

But what if we start this reaction with 10 moles of N_2(g) and 30 moles of H_2(g), how much energy would be released then?

If 1 mole of N_2(g) produced 92.4 kJ of energy, then 10 times as much N_2(g) (10 moles of N_2(g)) should produce 10 times as much energy:

10N_2(g) + 30H_2(g) → 20NH_3(g)   energy = 10 × ΔH

But what if we use 10 moles of H_2(g) and an excess of N_2(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 H_2(g).
We can do this by dividing the entire chemical equation by the stoichiometric coefficient (mole ratio) of H_2(g) in the chemical equation, that is, divide every term in the chemical equation (including the value of ΔH!) by 3:

From this new balanced chemical equation we see that when 1 mole of H_2(g) is consumed (with excess N_2(g)) to produce ²/₃ moles of NH_3(g), 30.8 kJ of energy will be released.

If 1 mole H_2(g) produced 30.8 kJ of energy, then 10 times the amount of H_2(g) will produce 10 times as much energy:

energy released = 10 mol × 30.8 kJ mol^-1 = 308 kJ

If we think about what we have done here we have:

1. divided ΔH value by the stoichiometric coefficient of H_2(g) in the balanced chemical equation
2. multiplied this new value by the moles of H_2(g) actually present

We can generalise this for any balanced chemical equation:

aA + bB → cC + dD     ΔH = +h kJ mol^-1

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

1. divide ΔH value by a
   energy absorbed per mole of A = ΔH/a     kJ mol^-1
2. 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:

1. divide ΔH value by b
   energy absorbed per mole of B = ΔH/b     kJ mol^-1
2. 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:

1. divide ΔH value by c
   energy absorbed per mole of C = ΔH/c     kJ mol^-1
2. 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:

1. divide ΔH value by d
   energy absorbed per mole of D = ΔH/d     kJ mol^-1
2. 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):

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

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