When 50.0 g of nitrogen react with excess hydrogen to form ammonia gas, 164.5 kJ of heat are liberated (released) at standard state conditions. Calculate the standard enthalpy of formation (in kJ/mol) for ammonia gas.

THE STANDARD ENTHALPY OF FORMATION OF AMMONIA GAS IS 293.75kJ OF HEAT.

Explanation:

To solve this question, you must first write out the equation for the reaction.

Equation:

N2 (g) + 3H2(g) <> 2NH3(g)

So therefore, when 50 g of N2 reacts, 164.5 kJ of Heat was liberated.

First equate the number of moles of Nitrogen and ammonia gas

1 mole of N2 produces 2 moles of ammonia

Calculate the molar mass of each variables:

Molar mass of N2 = 14*2 = 28 g/mol

Molar mass of ammonia = ( 14 + 1*3) = 17 g/mol

So, 1 mole of N2 = 2 moles of NH3

28 g/mol of N2 = 17 * 2 g/mol of NH3

If 50 g of nitrogen was used to react with excess hydrogen, the mass of ammonia formed is;

28 g of N2 = 34 g/mol of NH3

50 g of N2 = ( 50 * 34 / 28 ) g of NH3

= 1700 / 28

= 60 .71 g of ammonia.

At standard conditions, 34 g of ammonia will liberate 164.5 kJ of heat. What amonut would be generated by 60.71 g of ammonia?

34 g of ammonia = 164.5 kJ of heat

60.71 g of ammonia = ( 60.71 * 164.5  / 34) kJ of heat

= 9987.5 / 34

= 293.75 kJ of heat.

In other words, the standard enthalpy of formulation for ammonia gas is 293.75 kJ of heat.

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