Question

the temperature was recorded for 100 mL solution of (55 mL of 1.0 M HCL + 45 mL H2O) placed in a calorimeter before and after adding the NaOH (s). average temp. before mixing NaOH (s): 22.96 4th min: mixed 5th-12 min: 24.3, 25.0, 25.4, 25.6, 25.7, 25.8, 25.9, 25.9 mass of NaOH(s): 0.424 g what is the heat gained by the solution? Heat of the reaction? enthalpy of reaction? measured based on 0.5 M of solution specific heat=4.0 j/gC the density=1.02 g/mL

Answer 1

Q = m*Cp*dT

Hrxn = Q/n

V = 100 ml an dD = 1.02

then

m = 100*1.02 = 102 g

Cp = 4 J/gC

dT = Tf-Ti = 25.9 - 22.96 = 2.94 °C (choose the highest T)

then

Q = 102*4*(2.94) = 1199.52 J

Heat of reaction = Q = 1199.52 J (heat gained by solution)

for HRxn

we need mol of either base or acid ( the limiting reactant)

mol of HCl = M*V = 55*1 = 0.055 mmol of HCl

mol of NaOH = mass/MW = 0.424/40 = 0.0106 mol of NaOH

NaOH is limiting, so choose 0.0106

Then

HRxn = Q/n = 1199.52 /0.0106 = 113162.264151 J

Enthalpy of Reaction = -13162.264151 J (negative since exothermic)