Question

At 25°C, the standard enthalpy of combustion of gaseous propane (C3H8) is –2219.0 kJ per mole of propane, and the standard enthalpy of combustion of gaseous propylene (C3H6) is –2058.3 kJ per mole of propylene.
What is the standard enthalpy change for the following reaction at 25°C? C3H6(g) + H2(g) → C3H8(g)Substance∆H°f (kJ/mol)CO2(g)–393.5H2O(l)–285.8

Answer 1

Answer: The standard enthalpy change for the given reaction is 868.05 kJ

Explanation:

• Calculating the enthalpy of propane:

The chemical equation for the combustion of propane follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(C_3H_8(g))})+(5\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=-2219.0kJ$

Putting values in above equation, we get:

$-2219.0=[(3\times (-393.5))+(4\times (-285.8))]-[(1\times \Delta H^o_f_{(C_3H_8(g))})+(5\times (0))]\\\\\Delta H^o_f_{(C_3H_8(g))}=-140.7kJ/mol$

The enthalpy of formation of $C_3H_8$ is -140.7 kJ/mol

• Calculating the enthalpy of propylene:

The chemical equation for the combustion of propane follows:

$2C_3H_6(g)+9O_2(g)\rightarrow 6CO_2(g)+6H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(6\times \Delta H^o_f_{(CO_2(g))})+(6\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(C_3H_6(g))})+(9\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=-2058.3kJ$

Putting values in above equation, we get:

$-2058.3=[(6\times (-393.5))+(6\times (-285.8))]-[(2\times \Delta H^o_f_{(C_3H_6(g))})+(9\times (0))]\\\\\Delta H^o_f_{(C_3H_6(g))}=-1008.75kJ/mol$

The enthalpy of formation of $C_3H_6$ is -1008.75 kJ/mol

• Calculating the enthalpy change of the reaction:

The given chemical equation follows:

$C_3H_6(g)+H_2(g)\rightarrow C_3H_8(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_3H_8(g))})]-[(1\times \Delta H^o_f_{(C_3H_6(g))})+(1\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(C_3H_8(g))}=-140.7kJ/mol\\\Delta H^o_f_{(H_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_3H_6(g))}=-1008.75kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-140.7))]-[(1\times (-1008.75))+(1\times (0))]\\\\\Delta H^o_{rxn}=868.05kJ$

Hence, the standard enthalpy change for the given reaction is 868.05 kJ