Help with Year 12 thermo/chem?

2018-09-06 10:37 pm
Will be looking out for best answer. thanks

1. Given the following thermochemical equations, calculate the standard enthalpy of formation (in kilojoules per mole) of CuO(s).
2Cu(s) + S(s) Cu2S(s) [ΔrH° = –79.5 kJ/mol
S(s) + O2(g) SO2(g) [ΔrH° = –297 kJ/mol
Cu2S(s) + 2O2(g) 2CuO(s) + SO2(g) [ΔrH° = –527.5 kJ/mol

2. How much heat in kilojoules is needed to bring 2.48 kg of water from 36.6 to 98.0°C (comparable to making four cups of coffee)?

3. Nitric acid, HNO3, reacts with sodium hydroxide, NaOH , as follows:

HNO3(aq) + NaOH (aq) NaNO 3(aq) + H2O(l)
A student placed 12.5 mL of 1.4 M HNO3 in a coffee cup calorimeter, noted that the temperature was 24.8 °C, and added 43.7 mL of 1.3 M NaOH , also at 24.8 °C. The mixture was stirred quickly with a thermometer, and its temperature rose to 29.0 °C. Calculate the enthalpy of reaction in joules. Assume that the specific heats of all solutions are 4.18 J g-1 K-1 and that all densities are 1.00 g mL-1. Calculate the enthalpy of reaction per mole of acid (in units of kJ mol-1).

4. The addition of 335 J to 24.1 g of copper initially at 22°C will change its temperature to what final value (in °C). The specific heat of copper is 0.387 J g-1 K-1.


5. Which have a positive entropy change?


Petrol vaporises in the carburettor of a car engine.


Frost forms on the windscreen of your car.


Sugar dissolves in coffee.


Air is pumped into a tyre.


Moisture condenses on the outside of a cold glass.


Raindrops form in a cloud.

回答 (2)

2018-11-29 8:14 am
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2018-09-06 11:00 pm
1.
The equation for the standard enthalpy of formation is:
2 Cu(s) + O2(g) → 2 CuO(s)
So you need "2 Cu(s)" on the left. The only given equation that mentions Cu(s) is the first one, so copy the first given equation:
2 Cu(s) + S(s) → Cu2S(s) [ΔH° = –79.5 kJ/mol]
You also need "2 CuO(s)" on the right. The only given equation that mentions CuO(s) is the third one, so copy the third given equation:
Cu2S(s) + 2 O2(g) → 2 CuO(s) + SO2(g) [ΔH° = –527.5 kJ/mol]
Add the two equations here:
2 Cu(s) + S(s) + Cu2S(s) + 2 O2(g) →
Cu2S(s) + 2 CuO(s) + SO2(g) [ΔH° = –79.5 kJ/mol –527.5 kJ/mol]
Cancel like amounts on opposite sides of the arrow, and do the arithmetic for ΔH°:
2 Cu(s) + S(s) + 2 O2(g) → 2 CuO(s) + SO2(g) [ΔH° = –607.0 kJ/mol]
Now you need to get rid of the S(s) on the left and the SO2(g) on the right. Subtracting the second given equation will do that:
2 Cu(s) + S(s) + 2 O2(g) – S(s) – O2(g) →
2 CuO(s) + SO2(g) – SO2(g) [ΔH° = –607.0 kJ/mol – (–297 kJ/mol)]
Do the arithmetic everywhere:
2 Cu(s) + O2(g) → 2 CuO(s) [ΔH° = –310 kJ / mol]


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