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What is the total enthalpy change/ Hess Law?
H2S04(aq) + 2NaOH(aq) ---> Na2SO4 (aq) + 2H20(l)
If given an enthalpy value for this reaction, to be used in a Hess cycle is given, lets just say of -50KJ mol.
Then is it necessary to multiply this by 2, due to 2 moles of NaOH forming 2 mols H20?
Or is it counted as 1 enthalpy change?
Thanks
If given an enthalpy value for this reaction, to be used in a Hess cycle is given, lets just say of -50KJ mol.
Then is it necessary to multiply this by 2, due to 2 moles of NaOH forming 2 mols H20?
Or is it counted as 1 enthalpy change?
Thanks
ViolinGirl
H2S04(aq) + 2NaOH(aq) ---> Na2SO4 (aq) + 2H20(l)
If given an enthalpy value for this reaction, to be used in a Hess cycle is given, lets just say of -50KJ mol.
Then is it necessary to multiply this by 2, due to 2 moles of NaOH forming 2 mols H20?
Or is it counted as 1 enthalpy change?
Thanks
If given an enthalpy value for this reaction, to be used in a Hess cycle is given, lets just say of -50KJ mol.
Then is it necessary to multiply this by 2, due to 2 moles of NaOH forming 2 mols H20?
Or is it counted as 1 enthalpy change?
Thanks
∆H for neutralization is expressed in kJ/mol of water(ie combined volume of both the H2SO4 and NaOH used, take volume in cm^3 = mass in g as density of water is 1 g/cm cube)
Personally, I don't think you'd have to multiply by any mole ratio. I could be wrong on this.
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