When 4.5g of liquid water was formed by burning hydrogen gas in oxygen gas calculate the standard of heat formation of water?
The standard heat of formation of water is the amount of heat released or absorbed when one mole of water is formed from its elements (hydrogen and oxygen) at standard conditions (25°C and 1 atm pressure).
The balanced chemical equation for the combustion of hydrogen gas in oxygen gas to form water is:
2H2(g) + O2(g) → 2H2O(l)
From this equation, we can see that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of liquid water.
Given that 4.5g of water was formed, we can calculate the number of moles of water produced as follows:
moles of water = mass/molar mass = 4.5g / 18.015 g/mol = 0.2499 mol (rounded to 4 significant figures)
Since two moles of water are formed for every two moles of hydrogen gas and one mole of oxygen gas, the amount of moles of hydrogen gas consumed in the reaction is:
moles of H2 = 0.2499 mol / 2 = 0.12495 mol (rounded to 5 significant figures)
And the amount of moles of oxygen gas consumed is:
moles of O2 = 0.12495 mol / 0.5 = 0.2499 mol (rounded to 4 significant figures)
We can now use the standard heat of formation values for hydrogen gas, oxygen gas, and liquid water to calculate the standard heat of formation of water:
ΔHf° = (2 × ΔHf°(H2O) - ΔHf°(H2) - 0.5 × ΔHf°(O2))
where:
ΔHf°(H2) = 0 kJ/mol (standard heat of formation of hydrogen gas)
ΔHf°(O2) = 0 kJ/mol (standard heat of formation of oxygen gas)
ΔHf°(H2O) = -285.8 kJ/mol (standard heat of formation of liquid water)
Substituting these values into the equation, we get:
ΔHf° = (2 × (-285.8 kJ/mol) - 0 kJ/mol - 0.5 × 0 kJ/mol) = -571.6 kJ/mol
Therefore, the standard heat of formation of water is -571.6 kJ/mol, which means that this amount of heat is released when one mole of liquid water is formed from its elements (hydrogen and oxygen) under standard conditions.