A 50.0-mL volume of 0.400 M HCIO_4 at 24.35degreeC is added to 50.0 mL of 0.400

Question

A 50.0-mL volume of 0.400 M HCIO_4 at 24.35degreeC is added to 50.0 mL of 0.400 M KOH, also at 24.35degreeC. The final temperature is 27.06degreeC. What is the enthalpy change, deltaH_rxn, for the reaction, in kilojoules per mole? Assume that the solution has the same density and heat capacity as water. HCI0_4(aq) + KOH_(aq) rightarrow KCIO_4(aq) + H_20(l) deltaH_rxn = ? -1.13 kJ/mol -12.8 kJ/mol -27.4 kJ/mol -42.8 kJ/mol -56.7 kJ/mol In order to warm 1.5 L of tea from 22degreeC to 45degreeC, a cook places a 500.g block of heated stone into the teapot. What was the temperature of the stone before it was added to the tea? density of tea = 1.0 g/mL specific heat of tea = 4.184 J/gdegreeC specific heat of stone = 2.449 J/gdegreeC The initial temperature of the stone was: 68degreeC 84degreeC 118degreeC 140degreeC 163degreeC

Explanation / Answer

15) Total volume of the solution = 100 ml

Thus, total mass of the solution = volume*density = 100 g

Thus, heat evolved = mass of the solution*heat capacity*rise in temperature = 100*4.184*2.71 = 1133.86 J =1.134 kJ

Now, since the heat is evolved, delta H = -1.134 kJ

Now, moles of KOH reacted = moles of HClO4 reacted = molarity*volume of solution in litres = 0.4*0.05 = 0.02

Thus, delta H = -1.134/0.02 = -56.7 kJ/mole

2) Heat gained by tea = heat lost by stone

or, 1500*4.184*(45-22) = 500*2.449*(T-45)

or, T = 162.88 0C ; where T = initial temperature of the stone

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