what is the enthalphy change for the formation of 1.00 gram of waater vapour from hydrogen and oxygen
To calculate the enthalpy change for the formation of water vapor from hydrogen and oxygen, you need to use Hess's Law or enthalpy of formation data.
Hess's Law states that if a reaction can be expressed as the sum of several other reactions, the overall enthalpy change is the sum of the enthalpy changes of each individual reaction.
The enthalpy change for the formation of 1 mole of water (liquid) from its elements can be found in tables or literature. It is typically given as -285.8 kJ/mol.
Since we need to calculate the enthalpy change for the formation of 1.00 gram of water vapor, we need to convert grams to moles. The molar mass of water is approximately 18.015 g/mol, so 1.00 gram is equivalent to 1.00/18.015 = 0.0555 moles.
Now, using the stoichiometry of the balanced chemical equation for the formation of water vapor from hydrogen and oxygen (2H2 + O2 → 2H2O), we see that for the formation of 2 moles of water vapor, we need 2 moles of hydrogen and 1 mole of oxygen.
Therefore, for the formation of 0.0555 moles of water vapor, we need 0.0555/2 = 0.0277 moles of hydrogen and 0.0555/1 = 0.0555 moles of oxygen.
Finally, to calculate the enthalpy change for the formation of 0.0555 moles of water vapor, we use the enthalpy of formation of water (-285.8 kJ/mol) and the stoichiometric coefficients of the balanced equation.
Enthalpy change = (0.0555 moles) x (-285.8 kJ/mol) = -15.9 kJ
Therefore, the enthalpy change for the formation of 1.00 gram of water vapor from hydrogen and oxygen is approximately -15.9 kJ.