Question

What would be heat released when : 

(i) 0.25 mole of HCl in solution is neutralized 0.25 mole of NaOH solution? 

(ii) 0.5 mole of HNO₃ in solution is mixed with 0.2 mole of KOH solution? 

(iii) 200 cm³ of 0.2 M HCl is mixed with 300 cm³ of 0.1 M NaOH solution? 

(iv) 400 cm³ of 0.2 M H₂SO₄ is mixed with 300 cm³ of 0.1 M NaOH solution?

Answer 1

(i) H⁺(0.25mole) + OH^–(0.25mole) → H₂O(0.25mole) 

Heat released = 0.25 × 57.1 = 14.3 kJ 

(ii) 0.5 mole HNO₃ (aq) + 0.2 mole KOH(aq) 

The net reaction is (0.2 mole HNO₃ reacts with only 0.2 mole KOH) 0.2 mole is limiting reagent. 

0.2 mole of H⁺ + 0.2 mole of OH^– → 0.2 mole of H₂O 

0.3 mole of H⁺ will remain unreacted. 

Heat evolved = 0.2 × 57.1 = 11.4 kJ

(iii) 

0.03 mole of OH^– is limiting reactant 

Heat evolved = 0.03 × 57.1 = 1.71 kJ

(iv)

 

0.03 is limiting reagent Heat evolved = 0.03 × 57.1 = 1.7 kJ 

(v) Mass of solution = 200 + 300 = 500 g in (iii) part (Assuming d = 1 g cm³) 

q = m × c × T 1.71 × 1000 J = 500 g × 4.18 × T 

T = 0.82 K For (iv), T = \(\frac{1.7 \times 1000}{700 \times 4.18}\) = 0.58K