When 25.0 mL of 0.700 mol/L NaOH was mixed in a calorimeter with 25.0 mL of 0.700 mol/L HCl, both initially at 20.0 °C, the temperature increased to 22.1 °C. The heat capacity of the calorimeter is 279 J/°C. What is the enthalpy of neutralization in kJ/ mole of HCl?
Since the solutions are mostly water, the solutions are assumed to have a density of 1.0 g/mL and a specific heat of 4.18 J/g°C.
Select one:
a. −1020
b. -58.6
c. −5856
d. 58.6
no of moles of HCl = molarity * volume in L
= 0.7*0.025 = 0.0175 moles
total volume of solution = 25 +25 = 50ml
mass of solution = volume * desnity
= 50*1 = 50g
The heat involves are
heat from neutralization +heat to warm solution + heat to warm calorimeter = 0
q1 + q2 + q3 = 0
nH = mCT + CT = 0
0.0175mole*H = 50*4.18*(22.1-20) + 279*(22.1-20) = 0
H = -58560J/mole
= -58.56KJ/mole
round off -58.6KJ/mole
b. -58.6 >>>>answer
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