Calculate the molar enthalpy of reaction
standard enthalpy of formation below.
H20 = -285.8 kj/mole
H+ = 0.0 kj/mole
OH- = -229.9 kj/mol
H+(aq) + OH-(aq)→H2O(l)
For this, don't you do the summation of products x stoichemtry + the sum of reactants x stoich.
from there I'm not sure where to go.
Yes, that's what you do and when you get there there is nowhere else to go. You're subtracting kJ/mol from kJ/mol and the answer is in kJ/mol.
(1*-285.8)-(-229.9+0) = ?
Glycin heat at 200 degree centigrate
To calculate the molar enthalpy of reaction for the given chemical equation, you can use the standard enthalpies of formation of the compounds involved. The standard enthalpy of formation (ΔHf) is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.
In this case, you have the standard enthalpies of formation for H2O, H+, and OH-. The chemical equation provided is:
H+(aq) + OH-(aq) → H2O(l)
The stoichiometric coefficients of the reactants and products are both 1, indicating that one mole of each is involved in the reaction.
To calculate the molar enthalpy of reaction, you need to consider the difference in the standard enthalpies of formation between the products and the reactants. The enthalpy change of the reaction (ΔHrxn) can be determined using the following equation:
ΔHrxn = ∑(ΔHf(products) x stoichiometric coefficient) - ∑(ΔHf(reactants) x stoichiometric coefficient)
Let's apply this equation to calculate the molar enthalpy of reaction:
ΔHrxn = (ΔHf(H2O) x 1) - ((ΔHf(H+) x 1) + (ΔHf(OH-) x 1))
Substituting the given values:
ΔHrxn = (-285.8 kJ/mol x 1) - ((0.0 kJ/mol x 1) + (-229.9 kJ/mol x 1))
ΔHrxn = -285.8 kJ/mol + 0.0 kJ/mol + 229.9 kJ/mol
ΔHrxn = -55.9 kJ/mol
Therefore, the molar enthalpy of reaction for the given chemical equation is -55.9 kJ/mol.