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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 as:

NH_{3(g)} synthesis: N_{2(g)}+ 3H_{2(g)} → 2NH_{3(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:

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:

synthesis of ammonia: |
N_{2(g)} + 3H_{2(g)} → 2NH_{3(g)} |
ΔH = *-*92.4 kJ mol^{-1} |

decomposition of ammonia: |
2NH_{3(g)} → N_{2(g)} + 3H_{2(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}

## 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 (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)} + *3*H_{2(g)} → *2*NH_{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)} + *30*H_{2(g)} → *20*NH_{3(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 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 of H_{2(g)} in the chemical equation, that is, divide every term in the chemical equation (including the value of ΔH!) by 3:

original chemical equation |
N_{2(g)} |
+ |
3H_{2(g)} |
→ |
2NH_{3(g)} |
ΔH = -92.4 kJ mol^{-1} |

divide by 3 |
^{1}/_{3}N_{2(g)} |
+ |
^{3}/_{3}H_{2(g)} |
→ |
^{2}/_{3}NH_{3(g)} |
ΔH = -^{92.4}/_{3} kJ mol^{-1} |

^{1}/_{3}N_{2(g)} |
+ |
H_{2(g)} |
→ |
^{2}/_{3}NH_{3(g)} |
ΔH = -30.8 kJ mol^{-1} |

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 × 30.8 = 308 kJ

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

(a) divided ΔH value by the stoichiometric coefficient of H_{2(g)} in the balanced chemical equation

(b) multiplied this new value by the moles of H_{2(g)} actually present

We can generalise this for any balanced chemical equation:

*a**A* + *b**B* → *c**C* + *d**D* 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