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.

plasma is the state of matter where the substance is ionized and has no definite shape or volume. the fact of being ionized makes the substance a good electrical conductor.