Question

0.2 M, 100ml NaOH is mixed wih 0.4 M, 100mL HCl solution . Determine energy released during the reaction : Given H+(aq)OH−(aq)→H2o(l), ΔH=−57.5KJmol−1

Answer 1

Given : 0.2 M, 100ml NaOH is mixed wih 0.4 M, 100mL HCl solution.

and H⁺(aq) + OH¯ (aq) ⇔H₂O (l) , ∆H = -57.5 kJ/mol

To find : Determine the energy released during the reaction.

solution : no of moles of NaOH = 0.2 M × 100 ml

= 0.2 M × (100/1000) L

= 0.02 mol

similarly, no of moles of HCl = 0.4 M × 100 mL

= 0.4 M × (100/1000) L

= 0.04 mol

chemical reaction of sodium hydroxide and hydrochloride is given by,

NaOH + HCl ⇒NaCl + H₂O

one mole of NaOH reacts with one mole of HCl to produce one mole of NaCl and one mole water.

as, no of moles of HCl > no of moles of NaOH.

so NaOH is limiting reagent.

and hence, 0.02 mol of NaOH reacts 0.02 mol of HCl then form 0.02 mol of water and the same amount of NaCl.

so the energy released during reaction = ∆H × 0.02 mol

= -57 × 0.02 kJ

= -1.15 kJ

Therefore energy released during the reaction is 1.15 kJ.