HNO3 (g) + 4H2 (g) ---> 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?

You can get chemistry help at selfmath . webs . com

To calculate the enthalpy change when one mole of hydrogen reacts, we can use the balanced equation and the given enthalpy change:

HNO3 (g) + 4H2 (g) ---> NH3 (g) + 3H2O (g) ΔH = -637 kJ

From the equation, we see that 4 moles of hydrogen react to produce a change in enthalpy of -637 kJ. Therefore, dividing the enthalpy change by the coefficient of hydrogen, the enthalpy change when one mole of hydrogen reacts can be calculated as follows:

Enthalpy change when one mole of hydrogen reacts = (-637 kJ) / (4 mol)
= - 159.25 kJ/mol

Now, let's calculate the enthalpy change when 10.00 g of ammonia reacts with an excess of steam to form nitric acid and hydrogen gases.

First, we need to determine the number of moles of ammonia involved in the reaction. We can use the molar mass of ammonia (NH3) to convert grams to moles:

Molar mass of ammonia (NH3) = 14 g/mol + 3(1 g/mol) = 17 g/mol

Number of moles of ammonia = (10.00 g) / (17 g/mol)

Next, based on the balanced equation and the given enthalpy change, we know that the reaction produces 1 mole of HNO3 for every mole of NH3. Therefore, the enthalpy change for the reaction is the same as the enthalpy change for one mole of hydrogen reacting (from the previous calculation) since the reaction is balanced with respect to hydrogen.

So, the enthalpy change when 10.00 g of ammonia reacts with an excess of steam to form nitric acid and hydrogen gases is - 159.25 kJ/mol.