How much heat energy is given off when one kilogram of hydrogen gas at 25C and one atmosphere is combined with enough oxygen gas to make liquid water at 25C?
Write the equation.
Balance it.
dH = delta H.
dHrxn = (n*dHf products) - (n*dHf reactants)
I think I need more help. I tried but I think I'm going in circles on this one and not thinking clearly at this point.
Look in your text, usually in the back of the book, and find the delta H formation values for these compounds/elements at 25 C. I think the value for liquid water, although I'm not sure, is about -285 kJ/mol. The value for elements is arbitrarily assigned zero.
So 2H2(g) + O2(g) ==> 2H2O(l)
Then dHrxn = (2*dHf H2O) - (2*dHf H2 + dHf O2)
Solve for dH rxn and that will be for 1 mol H2. Then scale it up to 1000 g H2.
To determine the amount of heat energy given off when hydrogen gas reacts with oxygen gas to form liquid water, we need to calculate the enthalpy change or heat of reaction (∆H) for the reaction.
The balanced chemical equation for the reaction is:
2H₂(g) + O₂(g) → 2H₂O(l)
The enthalpy change (∆H) for this reaction is -286 kJ/mol, which means that when 2 moles of hydrogen gas react with 1 mole of oxygen gas to form 2 moles of liquid water, 286 kJ of heat energy is released.
To calculate the amount of heat energy released when one kilogram (1000 grams) of hydrogen gas reacts, we need to convert grams to moles and then multiply by the enthalpy change:
1 kg of hydrogen gas = 1000 grams
Molar mass of hydrogen (H₂) = 2 g/mol
Number of moles of hydrogen gas = (mass of hydrogen gas)/(molar mass of hydrogen)
= 1000 g / 2 g/mol
= 500 mol
Since 2 moles of H₂(l) release 286 kJ of heat energy, we can calculate how much heat energy is released by 500 moles:
Amount of heat energy released = (∆H) x (moles of H₂)
= -286 kJ/mol x 500 mol
= -143,000 kJ
Therefore, when one kilogram of hydrogen gas reacts with enough oxygen gas to form liquid water at 25°C and 1 atmosphere, approximately 143,000 kJ of heat energy is given off.