Thermochemical Equation

Question

When a sample of 134.28 g NH₄NO₃ dissolves in 55.01 g of water, the temperature changes from 25.72 C to 8.24 C. Calculate the heat of reaction in kJ/mol for this reaction (write answer to hundredths place).

Answer 1

NH₄NO₃ ==> NH₄⁺ + NO₃^-

q = mC∆T

q = heat = ?

m = mass of water = 55.01 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = 25.72º - 8.24º = 17.48º

q = (55.01 g)(4.184 J/gº)(17.48º) = 4023 J

134.28 g NH₄NO₃ x 1 mol NH₄NO₃/80.0 g = 1.68 moles NH₄NO₃

kJ/mol = 4023 kJ / 1.68 mols = 2395 J/mol = 2.40 kJ/mol