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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hey there!

boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey

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