Question

0.7180 ± 0.0002 g of KNO3 was added to 100.00 ± 0.02 g of H2O in a solution calorimeter. (a) (1 pt) Calculate ΔHrxn if Csystem = 121.46 ± 1.23 cal/°C and ΔT = -0.508 ± 0.009 °C. (b) (2 pts) Calculate the error in ΔHrxn. (c) (1 pt) Calculate the molar enthalpy change Hrxn ~ Δ for the dissolution of KNO3 in water. (d) (2 pts) Calculate in the error in Hrxn ~ Δ .

Answer 1

Mass of the system = (0.718 ± 0.0002) + (100 ± 0.02) g

mass of the system = (0.718 + 100) ± (square root of ((0.0002)^2 + (0.02)^2))

Mass = 100.718 ± 0.0200009

Q = m * Cs * dT

Q = (100.718 ± 0.0200009) * (121.46 ± 1.23) * (-0.508 ± 0.009)

here the errors are absolute errors we have to convert them into relative errors while doing mutiplication or division operations.

Relative error = absolute error/original value * 100

for mass = 0.0200009/100.718 * 100 = 0.0198%

for heat capacity = 1.23/121.46 * 100 = 1.0126%

for temperature = 0.009/0.508 * 100 = 1.7716%

Q = (100.718 ± 0.0198) * (121.46 ± 1.0126) * (-0.508 ± 1.7716)

Q = (100.718 * 121.46 * (-0.508)) ± (square root of ((0.0198)^2 + (1.0126)^2 + (1.7716)^2))

Q = -6214.47 ± 2.0406% J

dH rxn = -6217.47 ± 2.0406% J