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# Thermochemistry Tutorial Questions watch

I am given a table of mass, Cp and enthalpy f and c data. So ask if more info is needed.

3. When 2.25 mg of anthracene, C14H10(s), was burned in a bomb calorimeter the temperature rose by 1.35 K. Calculate the calorimeter constant (the heat capacity of the calorimeter). By how much will the temperature rise when 135 mg of phenol, C6H5OH(s), is burned in the calorimeter under the same conditions? enthalpy change of combustion for C14H10(s) = -7061 kJ mol-1, enthalpy change for combustion for C6H5OH(s) = -3054 kJ mol-1
2. (Original post by MatthewTG)

I am given a table of mass, Cp and enthalpy f and c data. So ask if more info is needed.

3. When 2.25 mg of anthracene, C14H10(s), was burned in a bomb calorimeter the temperature rose by 1.35 K. Calculate the calorimeter constant (the heat capacity of the calorimeter). By how much will the temperature rise when 135 mg of phenol, C6H5OH(s), is burned in the calorimeter under the same conditions? enthalpy change of combustion for C14H10(s) = -7061 kJ mol-1, enthalpy change for combustion for C6H5OH(s) = -3054 kJ mol-1
To work out the specific heat capacity you would use.

E = mcΔT

But for a calorimeter all of the individual masses and specific heat capacities are bundled up into one constant, called the heat capacity of the system

E = heat capacity x ΔT

where the heat capacity is the energy required to change the temperature of the whole system (in this case a bomb calorimeter) by 1ºC

heat capacity = energy input/ΔT

If you know the mass of a substance burned and its enthalpy of combustion you can work out how much energy it releases.

Then use the relationship above..
3. (Original post by charco)
To work out the specific heat capacity you would use.

E = mcΔT

But for a calorimeter all of the individual masses and specific heat capacities are bundled up into one constant, called the heat capacity of the system

E = heat capacity x ΔT

where the heat capacity is the energy required to change the temperature of the whole system (in this case a bomb calorimeter) by 1ºC

heat capacity = energy input/ΔT

If you know the mass of a substance burned and its enthalpy of combustion you can work out how much energy it releases.

Then use the relationship above..
by energy do you mean q in kj or the enthalphy change given in the question
4. (Original post by MatthewTG)
by energy do you mean q in kj or the enthalphy change given in the question
The enthalpy change in the question is the energy released per mole of compound combusted.

You can use this and the number of moles burned to find the actual energy, q, released.

Then with the temperature change you can find the heat capacity of the apparatus.
5. (Original post by charco)
The enthalpy change in the question is the energy released per mole of compound combusted.

You can use this and the number of moles burned to find the actual energy, q, released.

Then with the temperature change you can find the heat capacity of the apparatus.
Thank you that helped a lot.
6. (Original post by charco)
The enthalpy change in the question is the energy released per mole of compound combusted.

You can use this and the number of moles burned to find the actual energy, q, released.

Then with the temperature change you can find the heat capacity of the apparatus.
I would also like some help on question 6. (The weird square sumbols just Mean the delta triangle in enthalpy change symbol, the o is the theta for standard conditions). The U part confuses me.

6. Given the reactions (1) and (2) below, determine (all at 298 K)

(a) rHo and rUo for reaction (3)
(b) fHo for both HI(g) and H2O(g)

(1) H2(g) + I2(s) → 2 HI(g) rHo = +52.96 kJ mol-1
(2) 2 H2(g) + O2(g) → 2 H2O(g) rHo = -483.64 kJ mol-1
(3) 4 HI(g) + O2(g) → 2 I2(s) + 2 H2O(g)
7. (Original post by MatthewTG)
I would also like some help on question 6. (The weird square sumbols just Mean the delta triangle in enthalpy change symbol, the o is the theta for standard conditions). The U part confuses me.

6. Given the reactions (1) and (2) below, determine (all at 298 K)

(a) rHo and rUo for reaction (3)
(b) fHo for both HI(g) and H2O(g)

(1) H2(g) + I2(s) → 2 HI(g) rHo = +52.96 kJ mol-1
(2) 2 H2(g) + O2(g) → 2 H2O(g) rHo = -483.64 kJ mol-1
(3) 4 HI(g) + O2(g) → 2 I2(s) + 2 H2O(g)
It's just Hess' law.

double equation 1 and flip it round (do the same to the energy value)
add equation 2 (do the same to the energy)
This gives the enthalpy change of equation 3.

I suppose ΔU is the internal energy change

and

ΔH = ΔU + PΔV

assume molar quantities and standard conditions to calculate ΔU

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