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The standard enthalpy of formation of gaseous nitrogen dioxide is 33.2 kJ/mol, liquid water is -285.8 kJ/mol, aqueous nitric acid is -207 kJ/mol and the standard enthalpy of formation of gaseous nitrogen monoxide is 91.3 kJ/mol . Calculate the ΔHorxn for the following reaction:   3NO2(g) +  H2O(l)   →   2 HNO3(aq) + NO(g)

Question

The standard enthalpy of formation of gaseous nitrogen dioxide is 33.2 kJ/mol, liquid water is -285.8 kJ/mol, aqueous nitric acid is -207 kJ/mol and the standard enthalpy of formation of gaseous nitrogen monoxide is 91.3 kJ/mol . Calculate the ΔHorxn for the following reaction:   3NO2(g) +  H2O(l)   →   2 HNO3(aq) + NO(g)

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Solution

To calculate the ΔHrxn for the reaction, we need to use the formula:

ΔHrxn = Σ ΔHf(products) - Σ ΔHf(reactants)

The ΔHf values for the reactants and products are given in the problem.

For the reactants: 3NO2(g): 3 * 33.2 kJ/mol = 99.6 kJ/mol H2O(l): -285.8 kJ/mol

For the products: 2HNO3(aq): 2 * -207 kJ/mol = -414 kJ/mol NO(g): 91.3 kJ/mol

Substitute these values into the formula:

ΔHrxn = [(-414 kJ/mol + 91.3 kJ/mol) - (99.6 kJ/mol - 285.8 kJ/mol)] ΔHrxn = [-322.7 kJ/mol - (-186.2 kJ/mol)] ΔHrxn = -322.7 kJ/mol + 186.2 kJ/mol ΔHrxn = -136.5 kJ/mol

So, the standard enthalpy change for the reaction is -136.5 kJ/mol.

This problem has been solved

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