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Wize University Chemistry Textbook > Thermodynamics: Part 1: (0th law, 1st law, Calorimetry, Enthalpy)

Hess' Law Method

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Hess' Law

If a reaction is carried out in a series of steps, ΔH for the overall reaction can be found from the sum of the enthalpy changes of the individual steps.

This is because enthalpy (H) is a state function and when it changes, it does not depend on the pathway taken!
ΔHrxn = ΔH1 + ΔH2 + ΔH3 + …

Example:
What is the ΔH for the reaction?
CO(g) + ½ O2(g) → CO2(g)


C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ
2C(s) + O2(g) → 2CO(g) ΔH = -221.1 kJ

1◯  C(s)+O2(g)→CO2(g)=−393.5 KJ2◯2C(s)+O2(g)→2CO(g)=−221.1 KJ\text{\textcircled 1} \space \space C (s) + O_2 (g) → CO_2 (g) = -393.5 \space KJ\\\text{\textcircled 2} 2 C (s) + O_2 (g) → 2 CO (g) = -221.1 \space KJ

Equation 1 can stay the same (we see that it has CO2 on the right side of the equation and in the overall equation we want CO2 to be on the right side. The coefficient of CO2(g) does not need to be changed either).

For equation 2, we see that it has 2CO on the right. The overall equation has 1 CO on the left. This means that we would need to flip equation 2 (multiple deltaH by -1 and multiply equation 2 by 1/2. In total for equation 2 we will multiply it by -1/2.

1) C(s) + O2(g) --> CO2(g) deltaH= -390kJ
2) CO(g) --> C(s) +1/2O2(g) deltaH= +110.55 kJ

Now for the overall reaction, add 1) and 2):
CO(g) + 1/2O2(g) --> CO2(g) deltaH= -279.45 kJ

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Wize Concept
The two things you need to remember for this method are:
1) Whenever we want to look at a reaction in reverse, we must multiply ΔH by -1
2) Whenever we want to multiply a reaction by a coefficient, we must multiply the ΔH of that step by that same coefficient!



Practice: Calculating the Enthalpy of a Reaction Using Hess' Law

What is the ΔHrxn for the following reaction?

2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(g)


1) 2C + 3H2 --> C2H6 ΔH=-84.68kJ

2) C + O2 --> CO2 ΔH=-394kJ

3) H2 + 1/2O2 --> H2O ΔH=-286kJ



Given that the ΔfH0\Delta_fH^0 of ozone O3(g){\rm O}_{3(g)} is 142.67 kJ mol−1142.67\ kJ\ mol^{-1} and the reaction shown below, calculate the enthalpy for the unknown reaction
3/2 O2+I(aq)−→IO3(aq)−ΔH0=−202 kJ mol−1IO3(aq)−→I(aq)−+O3(g)ΔH0=???\begin{array}{ll} \rm 3/2\ O_2\quad+\quad I_{(aq)}^-\quad\overrightarrow{\hspace{1.5cm}}\quad IO_{3(aq)}^- &\hspace{1.5cm} \Delta H^0=-202\ kJ\ mol^{-1}\\[10pt] \rm IO_{3(aq)}^-\quad \overrightarrow{\hspace{1.5cm}}\quad I_{(aq)}^-\quad +\quad O_{3(g)} &\hspace{1.5cm} \Delta H^0=??? \end{array}


Extra Practice
Practice: Calculating the Enthalpy of a Reaction Using Hess' Law
What is the ΔHrxn for the following reaction?
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
Hess' Law: Enthalpy of reaction
Given that the ΔfH0\Delta_fH^0 of ozone O3(g){\rm O}_{3(g)} is 142.67 kJ mol−1142.67\ kJ\ mol^{-1} and the reaction shown below, calculate the enthalpy for the unknown reaction
3/2 O2+I(aq)−→IO3(aq)−ΔH0=−202 kJ mol−1IO3(aq)−→I(aq)−+O3(g)ΔH0=???\begin{array}{ll} \rm 3/2\ O_2\quad+\quad I_{(aq)}^-\quad\overrightarrow{\hspace{1.5cm}}\quad IO_{3(aq)}^- &\hspace{1.5cm} \Delta H^0=-202\ kJ\ mol^{-1}\\[10pt] \rm IO_{3(aq)}^-\quad \overrightarrow{\hspace{1.5cm}}\quad I_{(aq)}^-\quad +\quad O_{3(g)} &\hspace{1.5cm} \Delta H^0=??? \end{array}
Practice: Calculating the Enthalpy of a Reaction Using Hess' Law
What is the ΔHrxn for the following reaction?
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
Practice: Calculating the Enthalpy of a Reaction Using Hess' Law
What is the ΔHrxn for the following reaction?
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
Hess' Law: Enthalpy of reaction
What is the ΔHrxn for the following reaction?
2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(g)
1) 2C + 3H2 --> C2H6 ΔH=-84.68kJ
Hess' Law: Enthalpy of reaction
Given that the ΔfH0\Delta_fH^0 of ozone O3(g){\rm O}_{3(g)} is 142.67 kJ mol−1142.67\ kJ\ mol^{-1} and the reaction shown below, calculate the enthalpy for the unknown reaction
3/2 O2+I(aq)−→IO3(aq)−ΔH0=−202 kJ mol−1IO3(aq)−→I(aq)−+O3(g)ΔH0=???\begin{array}{ll} \rm 3/2\ O_2\quad+\quad I_{(aq)}^-\quad\overrightarrow{\hspace{1.5cm}}\quad IO_{3(aq)}^- &\hspace{1.5cm} \Delta H^0=-202\ kJ\ mol^{-1}\\[10pt] \rm IO_{3(aq)}^-\quad \overrightarrow{\hspace{1.5cm}}\quad I_{(aq)}^-\quad +\quad O_{3(g)} &\hspace{1.5cm} \Delta H^0=??? \end{array}
Enthalpy of Formation
Chlorine trifluoride is an interhalogen compound that reacts, usually explosively, with almost every known substance on earth, even relatively inert compounds, including glass, sand, concrete, and asbestos--which is a notorious flame-retardant material. It is a stronger oxidizing agent than O2, and is usually stored in metal canisters that are coated with F2 gas. Do not try to put out a fire caused by a ClF3 spill using water, because that will result in an explosive reaction that produces HF, a deadly gas. Trials involving ClF3 as war weapons were done during WWII, but handling, storing, and transporting this compound was more trouble than it was worth. Evil scientists need to determine the enthalpy of formation of ClF3, represented by this equation:
ClF(g)+ F2(g) → ClF3(g)       ΔH=?ClF_{\left(g\right)}+\ F_{2\left(g\right)}\ \rightarrow\ ClF_{3\left(g\right)}\ \ \ \ \ \ \ \Delta H=?
However, they were only able to determine the enthalpies of the following reactions:
Hess' Law: Enthalpy of formation
Given that the ΔfH0\Delta_fH^0 of ozone O3(g){\rm O}_{3(g)} is 142.67 kJ mol−1142.67\ kJ\ mol^{-1} and the reaction shown below, calculate the enthalpy for the unknown reaction
3/2 O2+I(aq)−→IO3(aq)−ΔH0=−202 kJ mol−1IO3(aq)−→I(aq)−+O3(g)ΔH0=???\begin{array}{ll} \rm 3/2\ O_2\quad+\quad I_{(aq)}^-\quad\overrightarrow{\hspace{1.5cm}}\quad IO_{3(aq)}^- &\hspace{1.5cm} \Delta H^0=-202\ kJ\ mol^{-1}\\[10pt] \rm IO_{3(aq)}^-\quad \overrightarrow{\hspace{1.5cm}}\quad I_{(aq)}^-\quad +\quad O_{3(g)} &\hspace{1.5cm} \Delta H^0=??? \end{array}
Calculate the heat of formation for Ethyne (C2H2) using the thermochemical information below.
2 C(s)+H2(g)→ C2H2(g)         ΔH0 = ???2\ C_{\left(s\right)}+H_{2\left(g\right)}\rightarrow\ C_2H_{2\left(g\right)}\ \ \ \ \ \ \ \ \ \Delta H^0\ =\ ???
C2H2+5/2O2→2 CO2+H2OΔH0=−1299.5kJC+O2→CO2−393.5kJH2+1/2O2→H2O−285.8kJ\begin{array}{lr} C_2H_2 + 5/2 O_2 \rightarrow 2\ CO_2 + H_2O & \Delta H^0 = -1299.5 kJ \\ C + O_2 \rightarrow CO_2 & -393.5 kJ \\ H_2 + 1/2 O_2 \rightarrow H_2O & -285.8 kJ \\ \end{array}
Thermochemistry: Enthalpy of formation
Calculate ΔHof of SO2(g) given the following:
S(s)+1.5O2(g)→SO3(g),ΔH=−395.8kJS_{(s)}+1.5O_{2(g)}\to SO_{3(g)}, \Delta H=-395.8kJ
2SO2(g)+O2(g)→2SO3(g),ΔH=−198.2kJ2SO_{2(g)}+O_{2(g)}\to 2SO_{3(g)}, \Delta H=-198.2kJ
Enthalpy of Formation
Calculate ΔHf\Delta H_f of SO2SO_2
S+1.5O2→SO3ΔH=−395.8kJ2SO2+O2→2SO3ΔH=−198.2kJ\begin{array}{}S+1.5 O_2\rightarrow SO_3\quad \Delta H=-395.8 kJ\\2SO_2+O_2\rightarrow 2SO_3\quad \Delta H=-198.2kJ\end{array}
Hess' Law: Enthalpy of reaction
Given that the ΔfH0\Delta_fH^0 of ozone O3(g){\rm O}_{3(g)} is 142.67 kJ mol−1142.67\ kJ\ mol^{-1} and the reaction shown below, calculate the enthalpy for the unknown reaction
3/2 O2+I(aq)−→IO3(aq)−ΔH0=−202 kJ mol−1IO3(aq)−→I(aq)−+O3(g)ΔH0=???\begin{array}{ll} \rm 3/2\ O_2\quad+\quad I_{(aq)}^-\quad\overrightarrow{\hspace{1.5cm}}\quad IO_{3(aq)}^- &\hspace{1.5cm} \Delta H^0=-202\ kJ\ mol^{-1}\\[10pt] \rm IO_{3(aq)}^-\quad \overrightarrow{\hspace{1.5cm}}\quad I_{(aq)}^-\quad +\quad O_{3(g)} &\hspace{1.5cm} \Delta H^0=??? \end{array}
Practice: Calculating the Enthalpy of a Reaction Using Hess' Law
What is the ΔHrxn for the following reaction?
2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(g)