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5.3 Enthalpy  (Page 7/25)

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Writing reaction equations for Δ H f °

Write the heat of formation reaction equations for:

(a) C 2 H 5 OH( l )

(b) Ca 3 (PO 4 ) 2 ( s )

Solution

Remembering that Δ H f ° reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:

(a) 2 C ( s , graphite ) + 3 H 2 ( g ) + 1 2 O 2 ( g ) ⟶ C 2 H 5 OH ( l )

(b) 3 Ca ( s ) + 1 2 P 4 ( s ) + 4 O 2 ( g ) ⟶ Ca 3 ( PO 4 ) 2 ( s )

Note: The standard state of carbon is graphite, and phosphorus exists as P 4 .

Check your learning

Write the heat of formation reaction equations for:

(a) C 2 H 5 OC 2 H 5 ( l )

(b) Na 2 CO 3 ( s )

Answer:

(a) 4 C ( s , graphite ) + 5 H 2 ( g ) + 1 2 O 2 ( g ) ⟶ C 2 H 5 OC 2 H 5 ( l ) ; (b) 2 Na ( s ) + C ( s , graphite ) + 3 2 O 2 ( g ) ⟶ Na 2 CO 3 ( s )

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Hess’s law

There are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment.

This type of calculation usually involves the use of Hess’s law    , which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps . Hess’s law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written:

C ( s ) + O 2 ( g ) ⟶ CO 2 ( g ) Δ H 298 ° = −394 kJ

In the two-step process, first carbon monoxide is formed:

C ( s ) + 1 2 O 2 ( g ) ⟶ CO ( g ) Δ H 298 ° = −111 kJ

Then, carbon monoxide reacts further to form carbon dioxide:

CO ( g ) + 1 2 O 2 ( g ) ⟶ CO 2 ( g ) Δ H 298 ° = −283 kJ

The equation describing the overall reaction is the sum of these two chemical changes:

Step 1: C ( s ) + 1 2 O 2 ( g ) ⟶ CO ( g ) Step 2: CO ( g ) + 1 2 O 2 ( g ) ⟶ CO 2 ( g ) ¯ Sum: C ( s ) + 1 2 O 2 ( g ) + CO ( g ) + 1 2 O 2 ( g ) ⟶ CO ( g ) + CO 2 ( g )

Because the CO produced in Step 1 is consumed in Step 2, the net change is:

C ( s ) + O 2 ( g ) ⟶ CO 2 ( g )

According to Hess’s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps. We can apply the data from the experimental enthalpies of combustion in [link] to find the enthalpy change of the entire reaction from its two steps:

C ( s ) + 1 2 O 2 ( g ) ⟶ CO ( g ) Δ H 298 ° = −111 kJ CO ( g ) + 1 2 O 2 ( g ) ⟶ CO 2 ( g ) C ( s ) + O 2 ( g ) ⟶ CO 2 ( g ) Δ H 298 ° = −283 kJ Δ H 298 ° = −394 kJ

The result is shown in [link] . We see that Δ H of the overall reaction is the same whether it occurs in one step or two. This finding (overall Δ H for the reaction = sum of Δ H values for reaction “steps” in the overall reaction) is true in general for chemical and physical processes.

A diagram is shown. A long arrow faces upward on the left with the phrase “H increasing.” A horizontal line at the bottom of the diagram is shown with the formula “C O subscript 2 (g)” below it. A horizontal line at the top of the diagram has the formulas “C (s) + O subscript 2 (g)” above it. The top and bottom lines are connected by a downward facing arrow with the value “Δ H = –394 k J” written beside it. Below and to the right of the top horizontal line is a second horizontal line with the equations “C O (g) + one half O subscript 2 (g)” above it. This line and the bottom line are connected by a downward facing arrow with the value “Δ H = –283 k J” written beside it. The same line and the top line are connected by a downward facing arrow with the value “Δ H = –111 k J” written beside it. There are three brackets to the right of the diagram. The first bracket runs from the top horizontal line to the second horizontal line. It is labeled, “Enthalpy of reactants.” The second bracket runs from the second horizontal line to the bottom horizontal line. It is labeled, “Enthalpy of products.” Both of these brackets are included in the third bracket which runs from the top to the bottom of the diagram. It is labeled, “Enthalpy change of exothermic reaction in 1 or 2 steps.”
The formation of CO 2 ( g ) from its elements can be thought of as occurring in two steps, which sum to the overall reaction, as described by Hess’s law. The horizontal blue lines represent enthalpies. For an exothermic process, the products are at lower enthalpy than are the reactants.

Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
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Muhammed
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specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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