Enthalpy change calculation

2007-10-19 9:05 pm

A student tried to determine the enthalpy change of the neturalization by putting 25.0cm3 of 1M HNO3 in a polystyrene cup and adding 25.0cm3 of 1M NH3 into it.The temperature rise recorded was 3.11"C.Given that the mass of the polystyrene cup is 250g,the S.H.C. of water and the polystyrene cup are 4200J/kg/K and 800J/kg/K respectively.Determine the enthalpy change of the neturalization.
(Density of water = 1 g/cm3)

回答 (1)

Carson

2007-10-19 11:33 pm

✔ 最佳答案

25.0cm^3 of 1M HNO3
25.0cm^3 of 1M NH3
polystyrene cup is 250g
S.H.C. of water and the polystyrene cup are 4200J/kg/K and 800J/kg/K respectively
The temperature rise recorded was 3.11 deg C
NH3 + H2O --> NH4OH
HNO3 + NH4OH --> NH4NO3 + H2O
mass of the 50cm^3 NH4NO3 is 50 * 1g = 50 g
mass of the polystyrene cup = 250 g
number of mole of H2O formed = (25 / 1000) * 1 = 0.025 mole
heat produced = 3.11 * 4200 * (50/1000) + 3.11 * 800 * (250/1000)
= 653.1 + 622 J
= 1275.1 J
the enthalpy change of the neturalization = 1275.1 J / 0.025 mole
= 51.004 kJ
I hope this can help your understanding. =)

參考: My knowledge

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