Question

**When 26.5 mL of 0.500 M
H_{2}SO_{4} is added to 26.5 mL of 1.00 M
KOH in a coffee-cup calorimeter at 23.50° C, the temperature rises
to 30.17° C. Calculate ΔH of this reaction. (Assume that
the total volume is the sum of the individual volumes and that the
density and specific heat capacity of the solution are the same as
for pure water.) (d for water = 1.00 g/mL; c for
water = 4.184 J/g°C).**

**Answer in kJ/molH _{2}O**

Answer #1

Total volume of water = 26.5 + 26.5 = 53.0 ml

Density of water = 1.00 g/ml

So, mass of water = volume*density = 53.0*1.00 = 53.0 g

mass = m = 53.0 g

specific heat = c = 4.184 J/g oC

Temperature rise = T = 30.17-23.50 = 6.67 oC

heat of reaction = H = -mcT

= -53.0*4.184*6.67 = -1479 J = -1.479 kJ

Moles of H2SO4 = concentration*volume = 0.500*0.0265 = 0.013 mol

Moles of KOH = 1.00*0.0265 = 0.026 mol

So, 0.013 mol of H2SO4 reacts with 0.026 mol of KOH to give 0.026 mol of H2O

H = -1.479/0.026 = -56.9 kJ/mol

When 23.8 mL of 0.500 M H2SO4 is added to 23.8 mL of 1.00 M KOH
in a coffee-cup calorimeter at 23.50°C, the temperature rises to
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volume is the sum of the individual volumes and that the density
and specific heat capacity of the solution are the same as for pure
water.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.)
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Δ
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Enter your answer in the provided box.
When 22.9 mL of 0.500 M H2SO4 is
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