Question

Say that 150 mL of a 1.50 M NaOH solution with a mass of 151 grams is added to 250 mL of a 1.00 M HCl solution with a mass of 252 grams in a coffee cup calorimeter. Before the mixing the temperature of each solution was 22 degrees celcius. After the reaction was complete, the temperature of the mixed solutions was 30.00 degrees celcius. The specific heat of the mixed solutions is essentially the same as water's, 41.184 J/g degrees celcius.

a.) What is the value of delta Hrxn?

b.) What is the balanced thermochemical equation for this reaction based on this data?

c.) Finally, calculate the delta Hrxn using delta Hf values and calculate the percent error for this experiment.

Answer 1

Tinitial = 22°C

Tfinal = 30°C

total mass of mix = 151 + 252 = 403 g of solution

a)

First, find total heat trnasferred

Q = m*C*(Tfinal - Tinitial)

C = 4.18 J/gC;

NOTE:  41.184 J/g degrees celcius. is not the acutal value, but 4.184 J/gC for WATER

Q = (403)(4.184)(30-22) = 13489.216 J

Total HRxn = Q = 13489.216 J = 13.489 kJ

HRXn rxn = -Q = -13.489 kJ (since this is exothermic, it is negative)

For HRxn per mol:

mol of NaOH = MV = 250*1 = 250 mmol = 0.25 mol

HRxn = -Q/(n)

HRxn = -13.489 kJ / (0.25) = 53.95 kJ/mol

b)

NaOH(aq) + HCl(aq) = H2O(l) + NaCl(aq) + heat

mol of NaOH = MV = 150*1.5 = 225 mmol = 0.225 mol of NaOH

mol of HCl = MV = 1*250= 250 mmol = 0.250 mol of HCl

limiting reactant is NaOH, so the calculation must be based on this

c)

HRXn = Hproducts- Hreactants

NaOH(aq) + HCl(aq) = H2O(l) + NaCl(aq) + heat

Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) = H2O(l) + Na+(aq) + Cl-(aq)

net ionic:

OH-(Aq) + H+(aq) = H2O(l)

HRxn = H2O(l) - (OH-(aq) + H+(aq))

HRxn = (-285.8) - (-230.0 + 0) = -55.8 kJ/mol

d)

% error for HRxn:

% Error = (55.8 - 53.95 ) / (53.95 ) *100% = 3.43%