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The following conventions apply when we use Δ H :

  1. Chemists use a thermochemical equation to represent the changes in both matter and energy. In a thermochemical equation, the enthalpy change of a reaction is shown as a Δ H value following the equation for the reaction. This Δ H value indicates the amount of heat associated with the reaction involving the number of moles of reactants and products as shown in the chemical equation . For example, consider this equation:

    H 2 ( g ) + 1 2 O 2 ( g ) H 2 O ( l ) Δ H = −286 kJ

    This equation indicates that when 1 mole of hydrogen gas and 1 2 mole of oxygen gas at some temperature and pressure change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released to the surroundings. If the coefficients of the chemical equation are multiplied by some factor, the enthalpy change must be multiplied by that same factor (Δ H is an extensive property):

    (two-fold increase in amounts) 2 H 2 ( g ) + O 2 ( g ) 2 H 2 O ( l ) Δ H = 2 × ( −286 kJ ) = −572 kJ ( two-fold decrease in amounts ) 1 2 H 2 ( g ) + 1 4 O 2 ( g ) 1 2 H 2 O ( l ) Δ H = 1 2 × ( −286 kJ ) = −143 kJ
  2. The enthalpy change of a reaction depends on the physical state of the reactants and products of the reaction (whether we have gases, liquids, solids, or aqueous solutions), so these must be shown. For example, when 1 mole of hydrogen gas and 1 2 mole of oxygen gas change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released. If gaseous water forms, only 242 kJ of heat are released.

    H 2 ( g ) + 1 2 O 2 ( g ) H 2 O ( g ) Δ H = −242 kJ
  3. A negative value of an enthalpy change, Δ H , indicates an exothermic reaction; a positive value of Δ H indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its Δ H is changed (a process that is endothermic in one direction is exothermic in the opposite direction).

Measurement of an enthalpy change

When 0.0500 mol of HCl( aq ) reacts with 0.0500 mol of NaOH( aq ) to form 0.0500 mol of NaCl( aq ), 2.9 kJ of heat are produced. What is Δ H , the enthalpy change, per mole of acid reacting, for the acid-base reaction run under the conditions described in [link] ?

HCl ( a q ) + NaOH ( a q ) NaCl ( a q ) + H 2 O ( l )

Solution

For the reaction of 0.0500 mol acid (HCl), q = −2.9 kJ. This ratio −2.9 kJ 0.0500 mol HCl can be used as a conversion factor to find the heat produced when 1 mole of HCl reacts:

Δ H = 1 mol HCl × −2.9 kJ 0.0500 mol HCl = −58 kJ

The enthalpy change when 1 mole of HCl reacts is −58 kJ. Since that is the number of moles in the chemical equation, we write the thermochemical equation as:

HCl ( a q ) + NaOH ( a q ) NaCl ( a q ) + H 2 O ( l ) Δ H = −58 kJ

Check your learning

When 1.34 g Zn( s ) reacts with 60.0 mL of 0.750 M HCl( aq ), 3.14 kJ of heat are produced. Determine the enthalpy change per mole of zinc reacting for the reaction:

Zn ( s ) + 2 HCl ( a q ) ZnCl 2 ( a q ) + H 2 ( g )

Answer:

Δ H = −153 kJ

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Be sure to take both stoichiometry and limiting reactants into account when determining the Δ H for a chemical reaction.

Another example of the measurement of an enthalpy change

A gummy bear contains 2.67 g sucrose, C 12 H 22 O 11 . When it reacts with 7.19 g potassium chlorate, KClO 3 , 43.7 kJ of heat are produced. Determine the enthalpy change for the reaction

C 12 H 22 O 11 ( a q ) + 8 KClO 3 ( a q ) 12 CO 2 ( g ) + 11 H 2 O ( l ) + 8 KCl ( a q ) .

Solution

We have 2.67 g × 1 mol 342.3 g = 0.00780 mol C 12 H 22 O 11 available, and 7.19 g × 1 mol 122.5 g = 0.0587 mol KClO 3 available. Since 0.0587 mol KClO 3 × 1 mol C 12 H 22 O 11 8 mol KClO 3 = 0.00734 mol C 12 H 22 O 11 is needed, C 12 H 22 O 11 is the excess reactant and KClO 3 is the limiting reactant.

The reaction uses 8 mol KClO 3 , and the conversion factor is 43.7 kJ 0.0587 mol KClO 3 , so we have Δ H = 8 mol × −43.7 kJ 0.0587 mol KClO 3 = −5960 kJ . The enthalpy change for this reaction is −5960 kJ, and the thermochemical equation is:

C 12 H 22 O 11 + 8 KClO 3 12 CO 2 + 11 H 2 O + 8 KCl Δ H = −5960 kJ

Check your learning

When 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of FeCl 2 ( s ) and 8.60 kJ of heat is produced. What is the enthalpy change for the reaction when 1 mole of FeCl 2 ( s ) is produced?

Answer:

Δ H = −338 kJ

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Questions & Answers

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cathode is a negative ion why is it that u said is negative
Michael Reply
cathode is a negative electrode while cation is a positive ion. cation move towards cathode plate.
king
CH3COOH +NaOH ,complete the equation
david Reply
compare and contrast the electrical conductivity of HCl and CH3cooH
Sa Reply
The must be in dissolved in water (aqueous). Electrical conductivity is measured in Siemens (s). HCl (aq) has higher conductivity, as it fully ionises (small portion of CH3COOH (aq) ionises) when dissolved in water. Thus, more free ions to carry charge.
Abdelkarim
HCl being an strong acid will fully ionize in water thus producing more mobile ions for electrical conduction than the carboxylic acid
Valentine
differiante between a weak and a strong acid
david
how can I tell when an acid is weak or Strong
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an aqueous solution of copper sulphate was electrolysed between graphite electrodes. state what was observed at the cathode
Bakanya Reply
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Enthalpy change of combustion: It is the enthalpy change when 1 mole of substance is combusted with excess oxygen under standard conditions. Elements are in their standard states. Conditions: pressure = 1 atm Temperature =25°C
Abdelkarim
Observation at Cathode: Cu metal deposit (pink/red solid).
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Equation at Anode: (SO4)^2- + 4H^+ + 2e^- __> SO2 + 2H2O
Abdelkarim
Equation : CuSO4 -> Cu^2+ + SO4^2- equation at katode: 2Cu^2+ + 4e -> 2Cu equation at anode: 2H2O -> 4H+ + O2 +4e at the anode which reacts is water because SO4 ^ 2- cannot be electrolyzed in the anode
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Covalent bonds are characterized by the sharing of electrons between two or more atoms. These bonds mostly occur between nonmetals or between two of the same (or similar) elements.
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covalent bonding is the mutual sharing of electrons between two element in a molecule, usually it involves non metals as they are less ionic and more electronegative than metals( ionic). and these bonds have high enthalpy of formation. and are strong bonds than most of the bond.
Chiranjeev
covalent bonding involves both nonmetals where there is complete sharing of electrons on the outermost energy level
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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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