Question

Complete the chemical and net ionic equations for the given reactions. Include the physical states of reactants and products if not given. Use the standard enthalpies of formation to calculate the changes in the standard enthalpies of reaction. (a) NaOH(aq) + HNO3(aq) — NaOH (aq) HO balanced net ionic equation: AH;xn =

Answer 1

NaOH (aq) + HNO3(aq) --> NaNO3 (aq) + H2O (l)

total ionic equation:
Na+(aq) + OH-(aq) + H+(aq) +NO3-(aq) ---> Na+(aq) + NO3-(aq) + H2O(
cancel spectator ions [Na+,NO3-]to get
Net ionic::

H+(aq) + OH-(aq) --> H2O (l)
****

dHrxn = dHf(NaNO3) + dHf(H2O) - dHf(NaOH) - dHf(HNO3)

take standard formation values from thermodynamic tables
dHrxn = [-446.2 Kj +-285.83 KJ]- [-470.114 kj]- [-207.36 kj]
= -54.556 KJ
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molecular equation
NH4NO3(aq) + NaOH(aq) ----> NH3(g) + NaNO3(aq) + H2O
Total ionic equation
NH4+ (aq) + Na+(aq) + OH-)(aq) ---> NH3(g) +H2O (l) + Na+(aq)
cancel spectator ions ; [Na+]to get

net ionic equation:
NH4+ (aq) + OH-(aq) ---> NH3(g) +H2O (l)

dHrxn = dHf(NaNO3) + dHf(H2O) +dHf(NH3) - dHf(NaOH) - dHf(NH4NO3)
=[-446.2 Kj +-285.83 KJ + [-46.1kj] -[-339.87 kj]-[-470.114kj]
= 31.854 kj
***
HNO3(aq) + NH3(aq) ---> NH4NO3(aq)

total ionic

H+(aq) + NO3-(aq) + NH3 (aq) ----> NH4+(aq) + NO3-(aq)

cancel spectator ions [NO3-] to get
net ionic equation:

H+(aq) + NH3 (aq) ----> NH4+(aq)

dHrxn = dHf(NH4NO3)- dHf(HNO3) -dH(NH3)
=[-339.87 KL]-[-207.36 KJ] -[-80.29 KJ]
= -52.22 KJ

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