Question

1. Using the standard enthalpies of formation, what is the standard enthalpy of reaction?

CO(g) + H₂O(g) --> CO₂(g) + H₂(g)

2. Calculate the standard enthalpy change for the following reaction at 25 �C.

HCl(g) + NaOH(s) --> NaCl(s) + H₂O(l)

3. For a particular isomer of C₈H₁₈, the following reaction produces 5113.3 kJ of heat per mole of C₈H₁₈(g) consumed, under standard conditions.

C₈H₁₈(g) + 25/2O₂(g) --> 8CO₂(g) + 9H₂O(g) (delta_rxn = -5113.3 kJ)

What is the standard enthalpy of formation of this isomer of C₈H₁₈(g)?

4. Many power plants produce energy by burning carbon-based fuels, which also produces carbon dioxide. Carbon dioxide is a greenhouse gas, so over-production can have negative effects on the environment. Use enthalpy of formation data to calculate the number of moles of CO₂(g) produced per megajoule of heat released from the combustion of each fuel under standard conditions (1 atm and 25 �C).

coal, C(s, graphite):

natural gas, CH₄(g):

propane, C₃H₈(g):

octane, C₈H₁₈(l) (?Hf� = �250.1 kJ �mol^-1):

Answer 1

Standard enthalpies of formation are used to find the standard enthalpy change of any reaction. Subtraction of the sum of the standard enthalpies of formation of the reactants (each being multiplied by its respective stoichiometric coefficient, ν) from the sum of the standard enthalpies of formation of the products (each also multiplied by its respective stoichiometric coefficient) gives standard enthalpy of reaction, ​

If the standard enthalpy of the products is less than the standard enthalpy of the reactants, the standard enthalpy of reaction will be negative. This implies that the reaction is exothermic. The converse is also true; the standard enthalpy of reaction will be positive for an endothermic reaction.

Elements in their standard states make no contribution to the enthalpy calculations for the reaction since the enthalpy of an element in its standard state is zero.

Answer for question (1):

Using the standard enthalpies of formation, what is the standard enthalpy of reaction?

CO(g) + H₂O(g) --> CO₂(g) + H₂(g)

standard enthalpies of formation of :

CO₂(g) = −393.509 kJ/mol

H₂(g) = 0

CO(g) = −110.525 kJ/mol

H₂O(g) = −241.818 kJ/mol

                                                                       CO(g)   +   H₂O(g) <==========>   CO₂(g)     +   H₂(g)

Standard enthalpies of formation of ​    −110.525    + (−241.818) <======>    −393.509    +   0

Net standard enthalpies of formation                    -352.343                                                −393.509

Standard enthalpy of reaction = Sum of standard enthalpies of formation of products - Sum of standard enthalpies of formation of reactants

= −393.509  - (- 352.343) = - 41.166 kJ/mol  

Standard enthalpy of reaction = - 41.166 kJ/mol  

The reaction is exothermic and spontaneous.

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Answer for question (2):

Calculate the standard enthalpy change for the following reaction at 25 ^oC.

HCl(g) + NaOH(s) --> NaCl(s) + H₂O(l)

standard enthalpies of formation of :

​ HCl(g) = −92.30 kJ/mol

NaOH(s) = −425.93 kJ/mol

NaCl(s) = −411.12 kJ/mol

H₂O(l) = −285.8 kJ/mol

standard enthalpy change = [(−411.12 + (−285.8)] - [−92.30 + (−425.93)]

                                           = [-696.92] - [- 518.23]

                                           = - 178.69 kJ/mol

Standard enthalpy of reaction = - 178.69 kJ/mol

The reaction is exothermic and spontaneous.

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Answer for question (3):

For a particular isomer of C₈H₁₈, the following reaction produces 5113.3 kJ of heat per mole of C₈H₁₈(g) consumed, under standard conditions.

C₈H₁₈(g) + 25/2O₂(g) --> 8CO₂(g) + 9H₂O(g) (delta_rxn = -5113.3 kJ)

What is the standard enthalpy of formation of this isomer of C₈H₁₈(g)?

standard enthalpies of formation of remaining reactants and products for the above reaction are as follows:=

C₈H₁₈(g) = ?

25/2O₂(g) = 0

8CO₂(g) = 8 x (−393.509 kJ/mol) = - 3148.072 kJ/mol

9H₂O(g) = 9 x (−241.818 kJ/mol) = - 2176.362 kJ/mol

Standard enthalpy of reaction = Sum of standard enthalpies of formation of products - Sum of standard enthalpies of formation of reactants

-5113.3 kJ = [- 3148.072 ​+ (- 2176.362)] - [Standard enthalpy of formation of C₈H₁₈(g)​]

-5113.3 kJ = [- 5324.434] - [Standard enthalpy of formation of C₈H₁₈(g)​]

[Standard enthalpy of formation of C₈H₁₈(g)​] = 5113.3 - 5324.434​ = - 211.34 kJ/mol

Standard enthalpy of formation of C₈H₁₈(g) = - 211.34 kJ/mol

So standard enthalpy of formation of C₈H₁₈(g) = -5113.3 kJ

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Answer for question (4):

Many power plants produce energy by burning carbon-based fuels, which also produces carbon dioxide. Carbon dioxide is a greenhouse gas, so over-production can have negative effects on the environment. Use enthalpy of formation data to calculate the number of moles of CO₂(g) produced per megajoule of heat released from the combustion of each fuel under standard conditions (1 atm and 25 ^oC).

coal, C(s, graphite):

natural gas, CH₄(g):

propane, C₃H₈(g):

octane, C₈H₁₈(l) (delta Hf = -250.1 kJ/mol^-1):

1 mega J = 1000 kJ

For coal, C(s, graphite) reaction is as follows:

C (s) + O2 (g)<========> CO2 (g)

Enthalpies of formation:

C = 0 kJ/mol

O2 = 0 kJ/mol

CO2 (g) = −393.509 kJ/mol

Standard enthalpy of reaction = Sum of standard enthalpies of formation of products - Sum of standard enthalpies of formation of reactants​

In the above reaction enthalpy of reaction equals to that of enthalpy of formation of CO2 product. So if 1 mega J enthalpy of reaction is produced then it equals to the total number of moles of CO2 produced in the reaction. Accordingly    

Since enthalpy of formation of CO2 (g) = −393.509 kJ/mol ​

So, −393.509 kJ/mol = Enthalpy of formation for 1 mole of CO2 (g)

then 1000 kJ will give how many moles of CO2 (g) = - 1000.0 / −393.509

                                                                               =  ​2.54 Mole of CO2 (g) produced.

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For natural gas, CH₄(g) reaction is as follows:

CH4 + 3/2 O2 <========> CO2 (g) + H2O (g)

Enthalpies of formation:

CH4 (g):= –74.6 kJ/mol

CO2 (g) = –393.5 kJ/mol

H2O (g) = –241.82 kJ/mol​

Standard enthalpy of reaction = Sum of standard enthalpies of formation of products - Sum of standard enthalpies of formation of reactants​

Standard enthalpy of reaction = [(–393.5) + (–241.82)] - [–74.6]

                                                = [-635.32] -   [–74.6]

                                                 = - 560.72 kJ/mol

For generation of 1 mega joule (1000 kJ) how many reaction runs have to be completed = 1000/560.72 = 1.783

So if 1 mega J (1000 kJ) enthalpy of reaction is to be produced then 1.78 moles of CH4 should undergo combustion to produce  1.783 of moles of CO2 and 1.783 of moles of H2O.

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For natural gas, C3H8 (g) reaction is as follows:

C3H8 (g) + 5O2 <========> 3CO2 (g) + 4H2O (g)

Enthalpies of formation:

C3H8 (g):= −104.6 kJ/mol

O2 = 0 kJ/mol

CO2 (g) = –393.5 kJ/mol x 3 = -1180.5 kJ/mol

H2O (g) = –241.82 kj/mol​ x 4= -967.28 kJ/mol

Standard enthalpy of reaction = Sum of standard enthalpies of formation of products - Sum of standard enthalpies of formation of reactants​

Standard enthalpy of reaction = [(-1180.5) + (-967.28)] - [−104.6]

                                                = [-2147.78] -   [–104.6]

Standard enthalpy of reaction = - 2043.18 kJ/mol

For generation of 1 mega joule (1000 kJ) how many reaction runs have to be completed = 1000/2043.18 = 0.489

So if 1 mega J (1000 kJ) enthalpy of reaction is to be produced then 0.489 moles of C3H8 should undergo combustion to produce (0.489x3=) 1.467 of moles of CO2 and (0.489x4=)1.956 moles of H2O.

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For octane, C₈H₁₈(l) reaction is as follows:

C8H18 (l)+ 25/2 O2 <========> 8CO2 (g)+ 9H2O (g)

Enthalpies of formation:

C8H18 (l):= −250.1 kJ/mol

O2 = 0 kJ/mol

CO2 (g) = –393.5 kJ/mol x 8 = -3148 kJ/mol

H2O (g) = –241.82 kj/mol​ x 9= - 2176.38 kJ/mol

Standard enthalpy of reaction = Sum of standard enthalpies of formation of products - Sum of standard enthalpies of formation of reactants​

Standard enthalpy of reaction = [(-3148) + (-2176.38)] - [−250.1]

                                                = [-5324.38] -   [–250.1]

Standard enthalpy of reaction = - 5074.28 kJ/mol

For generation of 1 mega joule (1000 kJ) how many reaction runs have to be completed = 1000/5074.28 = 0.197

So if 1 mega J (1000 kJ) enthalpy of reaction is to be produced then 0.197 moles of C8H18 should undergo combustion to produce (0.197x8=) 1.576 moles of CO2 and (0.197x9=)1.773 moles of H2O.