A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO

Question

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO_3 in a coffee cup calorimeter. If both solutions were initially at 35.00 degree C and the temperature of the resulting solution was recorded as 37.00 degree C determine the Delta H degree_rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HC1. Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2) that the density and the heat capacity of the resulting solution are the same as water.

Explanation / Answer

NaOH + HNO3 = H2O + NaNO3

mol of acid = MV = 0.3*100 = 30 mmol = 0.03 mol

mol of base = MV = 0.3*100 = 30 mmol

Tinitial = 35°C

Tfinal = 37°C

dT = 37-35 = 2 °C

exothermic

so

Qreaction = -Qwater

Qwater = m*C*(TF-Ti)

Qwater = (100+100)*4.184*(2) = 1673.6 J

Qrxn = -1673.6 J

HRxn = Qrxn/n = -1673.6 /0.03

HRxn= -55786.66 J/mol = -55.786 kJ/mol

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