Enthalpy of reaction can be found by using the sum of enthalpies of products minus the sum of reactants. Remember that the Heat of formation of elements in their standard states is 0 kJ/mol.
[2(-470.110kJ/mol+(0)]-[2(-285.8kJ/mol)+2(0)]= ?
Calculate the standard enthalpy of reaction for the reaction 2Na + 2H2O——> 2NaOH+ H2. Standard enthalpies of formation are -285.8 kJ/mol for H2O and -470.11 kJ/mol for NaOH.
3 answers
Right right, thank you so much. It's been a long day so my brain was in a bit of a mushy state. Thanks again!
eqn 1.....H2 + 1/2 O2 ==> H2O dH = -285.8 kJ/mol
eqn 2.....Na + 1/2 O2 + 1/2 H2 ==> NaOH dH = -470.11
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Multiply the reverse of equn 1 by 2 and add to twice eqn 2. Remember to change the sign of dH when a rxn is reversed. Add the two dH values.
Post your work if you get stuck. Check to make sure the equation at the end of the one you want.
eqn 2.....Na + 1/2 O2 + 1/2 H2 ==> NaOH dH = -470.11
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Multiply the reverse of equn 1 by 2 and add to twice eqn 2. Remember to change the sign of dH when a rxn is reversed. Add the two dH values.
Post your work if you get stuck. Check to make sure the equation at the end of the one you want.
1 answer