Question

2. When 2.76 g (0.0200 mol) of K2CO3 was mixed with 30.0 ml of approximately 2M HCl, the temperature rose by 5.2°C.

a) write a balanced equation for this reaction:

b) Calculate the enthalpy change of this reaction per mole of potassium carbonate. Assume that the specific heat of the final mixture is 4.184 J/g°C, and that its density is 1.00 g/ml.

3. When 2.00 g (0.0200 mole) of potassium hydrogen carbonate (KHCO3) is mixed with 30.0 ml of the same hydrochloric acid, the temperature falls by 3.7 °C.

a) write a balanced equation for this reaction.

b) Calculate the enthalpy change of this reaction per mole of KHCO3.

4. When KHCO3 is heated, it decomposes into potassium carbonate, water, and carbon dioxide. By applying Hess's law and using the results from #2 and #3, calculate the enthalpy change for the thermal decomposition of potassium hydrogen carbonate. Show all work and explain. Carefully note whether enthalpy changes are exothermic or endothermic.

Answer 1

2.

K₂CO₃ + 2 HCl $\to$ H₂O + CO₂  + 2 KCl. (1)

Heat change of solution (q_sol)

= Mass of solution × specific heat × temperature change

= 30.0 (mL) × 1.00(g/ml) × 4.184 × 5.2

= 653 J.

Moles of K₂CO₃ (n) = 0.0200 .

Hence,

Enthalpy change for this reaction

= - (q_sol/n)

= - ( 653 /0.0200)

= - 32650 J/mol

= - 32.65 KJ/mol.

Hence, enthalpy change is exothermic.

3.

KHCO₃ + HCl $\to$ KCl + H₂O + CO₂ (2)

Heat change of solution (q_sol)

= Mass × specific heat × temperature change

= 30.0 × 4.184 × (-3.7)

= - 464. J.

Moles of KHCO₃ (n) = 0.0200

Enthalpy of reaction

= - (q_sol/n)

= - (- 464/0.0200)

= - (-23200) J/mol

= + 23.20 KJ/mol.

Enthalpy change is endothermic.

4.

2 KHCO₃$\to$ K₂CO₃ + H₂O + CO₂  (3)

Eq.3 = 2× Eq.2 - Eq.1

Using hess law

Enthalpy change of Eq. 3

$\Delta$H₃ = 2×$\Delta$H₂ - $\Delta$ H₁

= 2× 23.20 - (-32.65)

= + 79.05 KJ

Enthalpy change of decomposition of KHCO₃ per mole

= + (79.05/2)

= + 39.525 KJ/mol.

Enthalpy change is endothermic.