Question

Calculate ΔH o rxn for the following reaction, after it is properly balanced with smallest whole-number coefficients: C2H6(g) + O2(g) → CO2(g) + H2O(g)[unbalanced] ΔH o f [C2H6(g)] = −84.667 kJ/mol ΔH o f [CO2(g)] = −393.5 kJ/mol ΔH o f [CO2(aq)] = −412.9 kJ/mol ΔH o f [H2O(g)] = −241.826 kJ/mol ΔH o f [H2O(l)] = −285.840 kJ/mol

Answer 1

Answer:

The enthalpy of the reaction is -2855.622 kilo Joules.

Explanation:

$2C_2H_6(g) + 7O_2(g)\rightarrow 4CO_2(g) + 6H_2O(g)$

We are given:

$\Delta H^o_f_{(C_{2}H_6(g))}= -84.667 kJ/mol$

$\Delta H^o_f_{O_2((g))}= 0 kJ/mol$

$\Delta H^o_f_{CO_2((g))}= -393.5 kJ/mol$

$\Delta H^o_f_{H_2O((g))}= -241.826 kJ/mol$

The equation used to calculate enthalpy of reaction :

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

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

$\Delta H^o_{rxn}=(4 mol\times \Delta H^o_f_{(CO_2(g))}+6 mol\times \Delta H^o_f_{(H_2O(g)))}-(2 mol\times \Delta G^o_f_{(C_{2}H_6(g))}+7 mol\times \Delta H^o_f_{(O_2(g)))$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[4 mol\times (-393.5 kJ/mol)+6 mol\times (-241.826 kJ/mol)]-[2 mol\times (-84.667 kJ/mol)+7 mol\times 0 kJ/mol]$

$=-2855.622 kJ$

The enthalpy of the reaction is -2855.622 kilo Joules.

Answer 2

The enthalpy change for the reaction is   -2855.622  kJ/mol

Recall that enthalpy(ΔH) is a state function so;

ΔHreaction = ∑ΔHproducts - ΔHreactants

So;

The equation of the reaction is; (Recall that the question specified that we should use the smallest whole number coefficients)

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

The enthalpy of each of the reactants and products are given below;

[C2H6(g)] = −84.667 kJ/mol

O2 g = 0 KJ/mol ( O2 exists in its standard state)

[CO2(g)]  =  −393.5 kJ/mol

[H2O(g)] = −241.826 kJ/mol

Hence;

ΔHreaction = ∑[4 × (−393.5) + 6 × (−241.826)] - [2 × (−84.667) + (7 × 0)]

ΔHreaction = -2855.622  kJ/mol

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