If 4.02g of hydrogen and 25.0g of oxygen (at standard conditions) are reacted to form water vapor, what quantity of heat is evolved from this reaction given that standard enthalpy of formation of H2O is -242kJ/mol?
A combination limiting reagent problem with a heat reaction problem.
2H2 + O2 ==> 2H2O
First let's determine the limiting reagent and the amount of H2O vapor formed.
mols H2 = g/molar mass = 4.02/2.01 = 2 mols.
mols O2 = 25/16 = 1.56
Now convert mols of each to mols H2O produced. For H2, it is
2.0 mols H2 x (2 mols H2O/2 mol H2) = 2 mols H2O vapor.
For O2 it is
1.56 mols O2 x (2 mols H2O/1 mol O2) = 3.12 mols H2O produced.
These answers are not the same; obviously one of them must be wrong. In limiting reagent problems the correct answer is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. So H2 is the limiting reagent and we will produce 2 mols H2.
Since the heat formation for H2O is 242 kJ/mol and we have two mols, the heat reaction will be 242 x 2 kJ.
Thank you Dr. Bob. Could you explain why you used double the molecular weight for hydrogen and not for oxygen? I thought that since both are diatomic that I would double it for both molecules.
Kay, I screwed up. Too much of a hurry I guess. You are absolutely correct. mols O2 = 25/32 = 0.781
That may change the limiting reagent so you need to go through the problem with that correction. Thanks for bringing this to my attention.
To calculate the quantity of heat evolved from a reaction, you can use the concept of stoichiometry and the given standard enthalpy of formation of water (H2O). Here are the steps to follow:
Step 1: Write and balance the chemical equation for the reaction.
The balanced chemical equation for the formation of water vapor from hydrogen and oxygen is:
2H2(g) + O2(g) -> 2H2O(g)
Step 2: Calculate the moles of reactants.
To do this, you need to convert the given masses of hydrogen and oxygen into moles.
- Hydrogen (H2): Given mass = 4.02 g. The molar mass of H2 is approximately 2 g/mol.
Moles of H2 = 4.02 g / 2 g/mol = 2.01 mol
- Oxygen (O2): Given mass = 25.0 g. The molar mass of O2 is approximately 32 g/mol.
Moles of O2 = 25.0 g / 32 g/mol = 0.78125 mol (approx.)
Step 3: Determine the limiting reactant.
Compare the moles of hydrogen and oxygen to determine the limiting reactant. The reactant that produces fewer moles of water is the limiting reactant.
From the balanced equation, we see that for every 2 moles of hydrogen, we need 1 mole of oxygen to form 2 moles of water. So, the limiting reactant will be oxygen because it produces fewer moles of water.
Step 4: Calculate the moles of water produced.
Since oxygen is the limiting reactant, we use its moles to find the moles of water produced.
From the balanced equation, we know that 1 mole of oxygen produces 2 moles of water. Therefore,
Moles of water = Moles of O2 x (2 moles of H2O / 1 mole of O2)
Moles of water = 0.78125 mol x (2 mol/1 mol) = 1.5625 mol
Step 5: Calculate the heat released using the standard enthalpy of formation.
From the given information, the standard enthalpy of formation of water vapor (H2O) is -242 kJ/mol.
To calculate the quantity of heat evolved, multiply the moles of water by the enthalpy change:
Heat evolved (in kJ) = Moles of water x Standard enthalpy of formation of H2O
Heat evolved = 1.5625 mol x (-242 kJ/mol)
Heat evolved = -377.1875 kJ
So, the quantity of heat evolved from this reaction is approximately -377.1875 kJ.
Note: The negative sign indicates that the reaction is exothermic (heat is evolved).