Question

1. Write balanced equations for the two reactions that will be carried out in this experiment HCL) + NaOH(aq) -----> HCION + NH3(04) -----> Also write a balanced equation for this reaction: NaOH + NH4Clao -->

Answer 1

1.

HCl is a strong acid and NaOH is a strong base. Hence, they will undergo neutralization reaction to form a salt and water as follows:

HCl(aq) + NaOH(aq) + NaCl(aq) + H200

Note that the reaction is already balanced as both sides have same number of each type of atoms.

HCl is a strong acid and NH3 is a weak Lewis base. Hence, they form a acid-base complex as given below:

HCl(aq) + NH3(aq) + NH4Cl(aq)

Note that the above reaction is also balanced.

Now, the balanced reaction between strong base NaOH and NH4Cl can be written as follows:

NaOH(aq) + NH4Cl (aq) + NaCl(aq) + NH3(aq) + H200

Note that this is simply a displacement reaction when cation of one reactant combines with anion of another.

2.

For a generic reaction

aA + bBC+ dD

The enthalpy change for the reaction,
HT
, can be written in terms of the enthalpies of formation, $\Delta H^\circ_{f}$ , of the reactants and products as follows:

ΔΗφανη = cx ΔΗ(C) + d x ΔΗ(D) – αX ΔΗ(Α) – 6x ΔΗ, (Β)

Where a, b, c and d are stoichiometric coefficients in the balanced reaction.

Hence, for our reactions:

HCl(aq) + NaOH(aq) + NaCl(aq) + H200

The enthalpy of the neutralization reaction $\Delta H_{x}$ can be written as

ΔΗ, = 1xΔΗ, (NaCl(aq) +1x ΔΗ; (Η2Ο) -1xΔΗ, (HCl(aq) - 1x ΔΗ(NaOH(aq). + ΔΗ = ΔΗ(NaClaq) + ΔΗ; (Η2Ο) – ΔΗ; (HClad)) – ΔΗ(NaOH(aq)).......... ....... (1)

For the second reaction:

HCl(aq) + NH3(aq) + NH4Cl(aq)

The enthalpy of the reaction $\Delta H_{y}$ can be written as

ΔΗ, = 1x ΔΗ (NH,CI (αα) – 1x ΔΗ (HCl(aq) - 1x ΔΗ (ΝΗ3(αα) = ΔΗ= ΔΗ(NH4Clag)) – ΔΗ; (HClag)) – ΔΗ(ΝΗ3(aq)).........(2)

Now, for the third reaction of NH4Cl and NaOH,

NaOH(aq) + NH4Cl (aq) + NaCl(aq) + NH3(aq) + H200

we can write the enthalpy change of the reaction as

ΔΗτανη = 1x ΔΗ(NaClaq) + 1x ΔΗ, (N H3(aq) +1 x ΔΗ, (Η2Ο) - 1x ΔΗ(NaOH (αα) - 1x ΔΗ(NH4Clag) Η ΔΗγκη = ΔΗ(NaCl αα)) + ΔΗ(ΝΗ3(αα)) + ΔΗ; (Η2Ο) – ΔΗ; (ΝαΟΗ (αα) – ΔΗ(NH4CIας)........(3)

Now, we have to prove that

ΔΗχη = ΔΗ - ΔΗ,

Hence, lets evaluate the right hand side of the equation and check if it equals the LHS.

Hence, subtracting equation (2 ) from equation (1).

ΔΗ-ΔΗ, = (ΔΗ (NaCl(aq) +ΔΗ, (Η2Ο) – ΔΗ; (HCl(aq)) - ΔΗ(ΝαΟΗ (aq)))- (ΔΗ (NH4Claq)) – ΔΗ;(HClaq)) - ΔΗ, (ΝΗ3(aq))) + ΔΗ-ΔΗ, = ΔΗ(NaCl(aq)) + ΔΗ; (Η2Ο) – ΔΗ(Head)) - ΔΗ(ΝαΟΗ (aq))- ΔΗ; (N H4Cl(aq) + ΔΗ(Hel(aq) + ΔΗ(ΝΗ3(aq) = ΔΗ,-ΔΗ, = ΔΗ; (NaCl(aq) + ΔΗ; (Η2Ο) – ΔΗ; (ΝαΟΗ (αα)-ΔΗ, (NH4Cl(aq) + ΔΗ(ΝΗ3(aq) + ΔΗ, – ΔΗ, = ΔΗτανη

Note that the right hand side of the third step of the above equation is equal to enthalpy change calculated in equation 3.

Hence, it is proved.