Calculate the expected change in temperature for 5.3 g of NH_4NO_3 dissolved in

Question

Calculate the expected change in temperature for 5.3 g of NH_4NO_3 dissolved in 80.0 mL of H_2O. Assume that the calorimeter absorbs no heat. (1 Mole NH_4NO_3 Delta H = -25.7 kJ/mol, 1 mol NH_4 NO_3 = 80.4g) q_solve = mC, Delta T Calculate the molar enthalpy of solution given the following Data. If we dissolve 3.0 g of salt in 100.0 mL of water, and the temperature went up 5.0 degree C. (1mL of water = 1g of water). Let's say it was done in a calorimeter that had a heat capacity of 13.4 J degree C^2. C_2 = 4.18 Jg C Salt M.W. = 35.0 g/mol -q_rxn = [mC, Delta t] + [C Delta t] Delta H = q_rxn/moles

Explanation / Answer

5. dH = q = mCpdT

with,

m = 80 ml x 1.00 g/ml = 80 g

Cp = 4.184 J/g.C

dT = ?

dH = 25.7 kJ/mol = (25.7 kJ x 1000 x 5.3 g/80.4 g/mol) J

we get change in temperature dT,

dT = 25.7 x 1000 x 5.3/80.4 x 4.184 x 80 = 5.061 oC

6. For the reaction,

-qrxn = qsoln + qcal

with,

qsoln = mCpdT = (100 ml x 1 g/ml) x 4.18 J/g.C x 5 oC = 2090 J

qcal = 13.4 x 5 = 67 J

-qrxn = 2090 + 67 = 2157 J

dH = 2157 J x 35 g/mol/3 g = 25.165 kJ/mol

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