HNO3 (g) + 4H2 (g) „³ NH3 (g) + 3H2O (g)deltaH = -637 kJ
Calculate the enthalpy change when one mole of hydrogen reacts and What is the enthalpy change when 10.00 g of ammonia is made to react with excess of steam to form nitric acid and hydrogen gases?
Answered above.
To calculate the enthalpy change for the given reaction, we can use the molar ratios and the given enthalpy change value.
1. Calculating the enthalpy change when one mole of hydrogen reacts:
- From the balanced equation, we see that 4 moles of hydrogen react to form 1 mole of ammonia.
- Therefore, the enthalpy change for one mole of hydrogen reacting can be calculated by dividing the given enthalpy change by 4:
Enthalpy change for one mole of H2 = -637 kJ / 4 = -159.25 kJ/mol
2. Calculating the enthalpy change when 10.00 g of ammonia is reacted with excess steam:
- First, we need to determine the number of moles of ammonia. To do this, we use the molar mass of ammonia:
Molar mass of NH3 = 14.01 g/mol (N) + 3(1.01 g/mol) (H) = 17.04 g/mol
Number of moles of NH3 = 10.00 g / 17.04 g/mol ≈ 0.586 mol (rounded to three decimal places)
- From the balanced equation, we see that 1 mole of ammonia forms 1 mole of nitric acid.
- Therefore, the enthalpy change for 0.586 moles of ammonia can be calculated by multiplying the enthalpy change for one mole of hydrogen with the ratio of ammonia to hydrogen moles:
Enthalpy change = (-159.25 kJ/mol) * 0.586 mol ≈ -93.256 kJ
In summary, the enthalpy change when one mole of hydrogen reacts is approximately -159.25 kJ/mol, and the enthalpy change when 10.00 g of ammonia is reacted with excess steam is approximately -93.256 kJ.