1. 25cm³ of 1.00mole/dm³ hydrochloric acid (HCl) at 21.5° placed in polystyrene cup and 25.0cm³ of 100mole/dm³ of NaOH at 21.5°C were added.
The mixed was stirred and the temperature rolls to 28.2°C
The density of each solution is 1.0gcm³ and the specific heat capacity of each solution is 4.18J/kg. Calculate the molar enthalpy of neutralization.
2. Calculate the heat of formation of methane given its heat of combustion as -811 Kg/mol and heats formation of C02 and H20 are -394 and -286K mol respectively.
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2018-01-29 10:08 pm
回答 (1)
2018-01-30 12:28 am
✔ 最佳答案
1. I'm going to assume that the NaOH solution was also 1.00 mol/dm^3. If that is not correct, you will need to do some additional things to determine the moles of NaOH used. Also, the units on your heat capacity should be 4.18 J/gC (not J/kg).If the two concentrations of NaOH and HCl are the same, then moles HCl = moles NaOH = 0.0250 dm^3 X 1.00 mol/L = 0.0250 mol
Heat absorbed by solution:
q = m c (T2-T1)
q = 50g (4.18 J/gC) (28.2 - 21.5) = 1.4X10^3 J
So, the neutralization reaction released -1.4X10^3 J
Molar enthalpy of neutralization = -1.4X10^3 J / 0.025 mol = -5.6X10^4 J/mol = -56 kJ/mol
2. Again on this question, your units are shown in correctly. Enthalpies of combustion and formation usually are express in kJ/mol.
The combustion of methane is shown by:
CH4(g) + 2O2(g) --> CO2(g) + 2 H2O(?) Delta Hcombustion = -811 kJ/mol
For this reaction:
Delta Hrxn = [DeltaHformation CO2 + 2 (DeltaHformation H2O)] - Delta Hformation CH4
-811 kJ/mol = [-394 - 2(286 kJ/mol)] - Delta Hformation CH4
Delta Hformation CH4 = +155 kJ/mol
收錄日期: 2021-04-24 00:54:58
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