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?
637 kJ/4 mols H2. So 637/4 = ?? for 1 mol H2.
For #2,
637 kJ/mol NH3 x [10 g NH3 x (1 mol NH3/17 g NH3)] = xx
thank Dr.Bob222
To calculate the enthalpy change when one mole of hydrogen reacts, we can use the given enthalpy change value of the reaction.
From the balanced equation:
HNO3 (g) + 4H2 (g) ---> NH3 (g) + 3H2O (g)
We see that 4 moles of hydrogen react to produce a change in enthalpy of -637 kJ.
So, the enthalpy change when one mole of hydrogen reacts can be calculated by dividing the given value by 4:
Enthalpy change per mole of hydrogen = -637 kJ / 4 = -159.25 kJ
Now, let's calculate the enthalpy change when 10.00 g of ammonia reacts with excess steam to form nitric acid and hydrogen gases.
We need to convert the given mass of ammonia (10.00 g) to moles. First, find the molar mass of ammonia (NH3):
1 mole of Nitrogen (N) = 14.01 g/mol
3 moles of Hydrogen (H) = 3 x 1.01 g/mol = 3.03 g/mol
Molar mass of NH3 = 14.01 g/mol + 3.03 g/mol = 17.04 g/mol
Now, calculate the number of moles of ammonia:
Number of moles = given mass / molar mass
Number of moles = 10.00 g / 17.04 g/mol ≈ 0.5866 mol
From the balanced equation, we can see that the reaction produces 1 mole of nitric acid (HNO3) and 4 moles of hydrogen (H2) for every mole of ammonia (NH3) that reacts.
Therefore, based on the stoichiometry of the reaction, we can say that when 0.5866 mol of ammonia reacts, it will produce 0.5866 mol of nitric acid and 4 x 0.5866 mol of hydrogen gas.
Now, we can use the enthalpy change value of the reaction to find the enthalpy change when 0.5866 mol of ammonia reacts:
Enthalpy change = number of moles x enthalpy change per mole
Enthalpy change for 0.5866 mol = 0.5866 mol x -637 kJ/mol = -373.58 kJ
So, the enthalpy change when 10.00 g of ammonia is made to react is approximately -373.58 kJ.
To calculate the enthalpy change when one mole of hydrogen reacts in the given reaction, we can use the stoichiometric coefficients and the given enthalpy change value.
First, we need to identify the stoichiometric coefficient of hydrogen in the balanced equation. In this case, the stoichiometric coefficient of hydrogen is 4.
Next, we can calculate the enthalpy change when one mole of hydrogen reacts by dividing the given enthalpy change by the stoichiometric coefficient of hydrogen:
Enthalpy change when one mole of hydrogen reacts = -637 kJ / 4 = -159.25 kJ
Therefore, the enthalpy change when one mole of hydrogen reacts is -159.25 kJ.
Now, to calculate the enthalpy change when 10.00 g of ammonia is made to react with an excess of steam, we need to follow these steps:
1. Calculate the moles of ammonia:
Molar mass of ammonia (NH3) = 14.01 g/mol + 3(1.01 g/mol) = 17.04 g/mol
Moles of ammonia = Mass of ammonia / Molar mass of ammonia
= 10.00 g / 17.04 g/mol
2. Convert the moles of ammonia to moles of nitric acid:
From the balanced equation, we know that 1 mole of ammonia reacts to form 1 mole of nitric acid. Therefore, the moles of nitric acid are equal to the moles of ammonia.
3. Calculate the enthalpy change using the stoichiometric coefficient of ammonia:
Enthalpy change = Moles of nitric acid × Enthalpy change when one mole of hydrogen reacts
So, let's calculate the moles of ammonia first:
Moles of ammonia = 10.00 g / 17.04 g/mol = 0.5877 mol (rounded to four decimal places)
Since we already know that the moles of nitric acid are equal to the moles of ammonia, we can proceed to calculate the enthalpy change:
Enthalpy change = 0.5877 mol × (-159.25 kJ/mol)
The enthalpy change when 10.00 g of ammonia is made to react with an excess of steam is determined by multiplying the moles of nitric acid by the enthalpy change when one mole of hydrogen reacts.
Therefore, let's calculate the final result:
Enthalpy change = 0.5877 mol × (-159.25 kJ/mol) = -93.52 kJ (rounded to two decimal places)
Therefore, the enthalpy change when 10.00 g of ammonia is made to react with an excess of steam is approximately -93.52 kJ.