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

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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