Question

I 1. The Heat of Neutralization of HCl(aq) and NaOH(aq) Mass Heat Capacity Initial Temp Solution 102g 4.18 J/gºC 20.00 C Final Temp 26.81 3-6 Initial Temp Final Temp nation: sely de 20.00 C 25.21C 2 The Heat of Solution for NaOH) Mass Heat Capacity Solution 2.00g 4.18 J/gºC 3 The Heat of Reaction for HClaq) and NaOH) Solution 2.00g 4.18 J/gºC The Heat of Reaction for HCl(aq) and NaOH) Solution 2.00g 4.18 J/g °C 20.00 31.83 INITIAL TEMP 20.00 C FINAL TEMP 31.61 c Analysis Demonstrate, by adding together the chemical equations, that the AH for part 4 should be the sum of the AH's you calculated in NO. 2 and 3. How closely does your calculated AH for part 4 compare to this sum? Find the percent difference/error. 1. The Heat of Neutralization of HCl(aq) and NaOH(aq) 2. The Heat of Solution for NaOH(s) 3. The Heat of Reaction for HCl(aq) and NaOH(s)

Answer 1

Answer-

Table for the 1st Reaction [Neutralization of HCl(aq.) by NaOH(aq.)] :-

	mass (m)	Heat capacity (s)	Initial temperature(T1)	Final temperature(T2)
Solution	102 gm	4.18 J /g°C	20°C	26.81°C

The equation for the neutralization of HCl(aq.) by NaOH(aq.) -

Na⁺ (aq) + OH⁺(aq) + H⁺ (aq) + Cl^-(aq) → H2O (l) + Na⁺ (aq) + Cl^- (aq)

For this reaction, the mass of the solution is = 102gm

Heat capacity of the solution = 4.18 J /g°C

Temperature change = T₂ - T₁ = 26.81°C - 20°C = 6.81°C

So, the calculation for heat released or absorbed(q) for the reaction-

q = mass x specific heat capacity x temperature change = m x s x$\Delta$t = 102 g x  (4.18 J /g°C) x 6.81°C

   q = 2903.5116 J

Enthalpy of the reaction is given by the formulae-

ΔH kJ / mol = q/1000 ÷ moles of the reactant

Now, since the reaction has 1:1:1 stoichiometry of both of the reactant and H₂O formed in the product

we can write moles of the reactant = mass of the solution / molar mass of H₂O = 102 / 18 = 6 mol

So, ΔH = 2903.5116 / 1000 ÷ 6 = 0.48 kJ / mol

Table for the 2nd Reaction [Dissolution of NaOH(s) ] :-

	mass (m)	Heat capacity (s)	Initial temperature(T1)	Final temperature(T2)
Solution	2.00 gm	4.18 J /g°C	20°C	25.21°C

The equation for the dissolution of NaOH(s)-

NaOH(s) $\rightarrow$ Na⁺(aq.) + OH^-(aq.)

For this reaction, the mass of the solution is = 2.00 gm

Heat capacity of the solution = 4.18 J /g°C

Temperature change = T₂ - T₁ = 25.21°C - 20°C = 5.21°C

So, the calculation for heat released or absorbed(q) for the reaction-

q = mass x specific heat capacity x temperature change = m x s x$\Delta$t = 2.00 g x  (4.18 J /g°C) x 5.21°C

   q = 43.5556 J

Now, enthalpy of the reaction is given by the formulae-

ΔH kJ / mol = q/1000 ÷ moles of reactant

Now, since the reaction has 1:1 stoichiometry of the reactant and H₂O formed in the product

we can write moles of the reactant = mass of the solution / molar mass of H₂O = 2.00 / 18 = 0.11 mol

Now, ΔH = 43.5556 /1000 ÷ 0.11 = 0.39 kJ / mol

Table for the 3rd Reaction [Reaction of HCl(aq.) and NaOH(s)] :-

	mass (m)	Heat capacity (s)	Initial temperature(T1)	Final temperature(T2)
Solution	2.00 gm	4.18 J /g°C	20°C	31.83°C

The equation for the reaction of HCl(aq.) by NaOH(s) -

NaOH (s) + H⁺ (aq) + Cl ^- (aq) → H2O (l) + Na⁺ (aq) + Cl ^- (aq)

For this reaction, the mass of the solution is = 2.00gm

Heat capacity of the solution = 4.18 J /g°C

Temperature change = T₂ - T₁ = 31.83°C - 20°C = 11.83°C

So, the calculation for heat released or absorbed(q) for the reaction-

q = mass x specific heat capacity x temperature change = m x s x$\Delta$t = 2.00 g x  (4.18 J /g°C) x 11.83°C

   q =  98.8988 J

Now, enthalpy of the reaction is given by the formulae-

ΔH kJ / mol = q/ no. of moles of reactant

Now, since the reaction has 1:1 stoichiometry of the reactant and H₂O formed in the product

we can write moles of the reactant = mass of the solution / molar mass of H₂O = 2.00 / 18 = 0.11 mol

Now, ΔH = 98.8988 /1000 ÷ 0.11 = 0.90 kJ / mol

Now, we will write the previous 3 chemical reaction with their respective $\Delta$ H in order-

Na⁺ (aq) + OH⁺(aq) + H⁺ (aq) + Cl^-(aq) → H2O (l) + Na⁺ (aq) + Cl^- (aq) ;  $\Delta$H_{neutraliazation} = 0.48 kJ / mol eqn- 1

NaOH(s) + $\rightarrow$ Na⁺(aq.) + OH^-(aq.) equation -2 ; $\Delta$ H_dissoulution =  0.39 kJ / mol equation - 2

NaOH (s) + H⁺ (aq) + Cl ^- (aq) → H2O (l) + Na⁺ (aq) + Cl ^- (aq) $\Delta$ H_reaction =  0.90 kJ / mol equation -3

Now, if we add equation 2 and 3 , we get-

Na⁺ (aq) + OH⁺(aq) + H⁺ (aq) + Cl^-(aq) + NaOH(s) + $\rightarrow$ H2O (l) + Na⁺ (aq) + Cl^- (aq) + Na⁺(aq.) + OH^-(aq.)

; $\Delta$ H = $\Delta$ H_{neutraliazation} +  $\Delta$H_dissoulution = 0.48 kJ + 0.39 kJ

If we cancel the common things which are on same side, we get-

NaOH (s) + H⁺ (aq) + Cl ^- (aq) → H2O (l) + Na⁺ (aq) + Cl ^- (aq) ; $\Delta$ H = 0.48 kJ + 0.39 kJ = 0.87 kJ

Thus, by adding reaction 1 and 2 we get reaction 3, and if we compare $\Delta$ _rxn of equation 3 with the sum of $\Delta$ Hs of the reaction 2 , we get almost same result

The % error between this 2 $\Delta$ H = [ (0.90 - 0.87) / 0.87 ] x 100 = 3.44%