Initial temp: 19.06 C trial 1: 0.5 M concentration trial 2 : 1.0 M concentration 200 mL of HCl and 200 mL of NaOH are combined in an insulated container. HCl = 0.5 mol/ml x 200 x 10^-3 ml = 0.1 mol NaOH = 0.5 mol/ml x 200 x 10^-3 ml= 0.1 mol H+(aq) + OH-(aq) - - -> H2O (l) ∆H H++ = 0, ∆H OH−− = -230 kJ/mol, ∆H H2O = -286 kJ/mol ∆H reactants = ∆H H++ + ∆H OH−− = -230 + 0 = -230 kJ/mol ∆H reactants = -286 kJ/mol - ( -230 kJ/mol) = -56 kJ/mol For each trial, calculate how much energy is released during this reaction.
Initial temp: 19.06 C
trial 1: 0.5 M concentration
trial 2 : 1.0 M concentration
200 mL of HCl and 200 mL of NaOH are combined in an insulated container.
HCl = 0.5 mol/ml x 200 x 10^-3 ml = 0.1 mol
NaOH = 0.5 mol/ml x 200 x 10^-3 ml= 0.1 mol
H+(aq) + OH-(aq) - - -> H2O (l)
∆H H++ = 0, ∆H OH−− = -230 kJ/mol, ∆H H2O = -286 kJ/mol
∆H reactants = ∆H H++ + ∆H OH−−
= -230 + 0
= -230 kJ/mol
∆H reactants = -286 kJ/mol - ( -230 kJ/mol)
= -56 kJ/mol
For each trial, calculate how much energy is released during this reaction.
trial 1: 0.5 M concentration
200 mL of HCl and 200 mL of NaOH are combined in an insulated container.
No of moles of HCl (n[HCl] ) = (0.5 mole/L) x (200 x 10-3 L) = 0.1 mole
No of moles of Na OH ( n[NaOH] ) = (0.5 mole/L) x (200 x 10-3 L) = 0.1 mole
0.1 H+(aq.) + 0.1 OH-(aq.) - - -> 0.1H2O (l)
∆H (H+) = 0,
∆H (OH)− = -230 kJ/mole,
∆H (H2O) = -286 kJ/mole
∆Hreactants = n[HCl] * ∆H (H+) + n[NaOH] * ∆H (OH)−
= (0.1 * 0)kJ/mole + 0.1 * -230 kJ/ mole
= -23 kJ/ mole
∆Hproducts= n[H2O] * ∆H (H2O)
= (0.1 * -286 )kJ/mole
= -28.6kJ/ mole
∆Hreaction = ∆Hproducts - ∆Hreactants
= [-28.6-(-23)] kJ/mole
= -5.6 kJ/mole
energy is released during this reaction (Trial 1) = -5.6 kJ/mole
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