Question

When 27.4 mL of 0.500 M H2SO4 is added to 27.4 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.) kJ/mol H2O

Answer 1

temprature changes from 23.5⁰c to 30.17⁰c therefore $\Delta$T = 30.17 - 23.5 = 6.67

total volume of solⁿ = volume of H₂SO₄ + volume of KOH =27.4+27.4=54.8 ml

density of solution is 1gram per ml therefore 54.8ml solution = 54.8 gram solution =mass of solution=54.8gram

specific heat of solution =specific heat of pure water = 4.184 J/g

we know formula $\Delta$H = mass of solution $\times$ specific heat $\times$$\Delta$T

substitute values in above equation

$\Delta$H = 54.8 $\times$ 4.184 $\times$ 6.67 = 1529.3 J

$\Delta$H for reactiion = 1529.3 J