In a coffee-cup calorimeter, 130.0 mL of 1.0 M NaOH and 130.0 mL of 1.0 M HCl are mixed. Both solutions were originally at 26.8°C. After the reaction, the final temperature is 33.5°C. Assuming that all the solutions have a density of 1.0 g/cm and a specific heat capacity of 4.18 J/°C ⋅ g, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.
Enthalpy change = kJ/mol
NaOH + HCl --> NaCl + H2O
each one mol of NaOH and HCl reacts with each other and produces 1 mol water.
moles water = 130 mL x 10-3 L x 1.0 M = 0.13 mol
total volume = 130 + 130 = 260 mL = 260 g (density of 1.0 g/cm3)
q = -mass x specific heat x temperature change
q = -260 g x 4.18 J/°C ⋅ g x (33.5 - 26.8)°C
q = -7281.56 J
molar enthalpy = -q / moles H2O = 7281.56 J / 0.13 mol = 56012 J/mol = 56.012 kJ/mol
molar enthalpy = -56.0 kJ/mol ( 3 significant figures) ---- please observe negative sign
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