The reaction through which carbon changes from graphite to diamond form is represented by the equation

C(graphite) ---> C(diamond) delta H: ?

Using the reactions below, determine the amount of heat absorbed in the change of one mole of graphite to one mole of diamond

C(graphite) + O2 ---> CO2(g) delta H is -393.5 kJ/mol

C(diamond) + O2(g) ---> CO2(g) Delta H is -395.4 kJ/mol

How much energy is absorbed in the manufacture of one kilogram of diamond?

If this reaction were to take place in water, how many grams of water could be cooled by 5 degrees Celsius by the formation of 1 mol of diamond

To determine the amount of heat absorbed in the change of one mole of graphite to one mole of diamond, we need to use the given reactions and their enthalpy values.

The reaction we are interested in is:
C(graphite) ---> C(diamond)

We can use the conservation of energy principle and the given reactions to solve for the enthalpy change of this reaction.

First, let's label the reactions:

1. C(graphite) + O2 ---> CO2(g) (Reaction A)
Enthalpy change: ΔH_A = -393.5 kJ/mol

2. C(diamond) + O2(g) ---> CO2(g) (Reaction B)
Enthalpy change: ΔH_B = -395.4 kJ/mol

Now, let's equate the two reactions to cancel out the common product CO2(g):

C(graphite) + O2 ---> CO2(g) (Reaction A)
C(diamond) + O2(g) ---> CO2(g) (Reaction B)

Subtracting Reaction A from Reaction B will give us the desired reaction:

C(diamond) - C(graphite) + O2(g) ---> 0
ΔH = ΔH_B - ΔH_A

Now, we can calculate the enthalpy change, ΔH, for the conversion of one mole of graphite to one mole of diamond:

ΔH = -395.4 kJ/mol - (-393.5 kJ/mol)
ΔH = -1.9 kJ/mol

This means that the reaction from graphite to diamond is exothermic, and it releases 1.9 kJ of energy per mole of graphite converted.

To determine the amount of heat absorbed in the manufacture of one kilogram of diamond, we need to convert the mass of diamond to moles. Since the molar mass of carbon is 12 g/mol, the mass of one mole of diamond would be 12 g.

1 kilogram = 1000 grams
Number of moles = mass (g) / molar mass (g/mol) = 1000 g / 12 g/mol ≈ 83.33 mol

Heat absorbed = ΔH * moles
Heat absorbed = -1.9 kJ/mol * 83.33 mol ≈ -158.33 kJ

Therefore, approximately -158.33 kJ of energy is absorbed in the manufacture of one kilogram of diamond.

For the second part of the question, to determine how many grams of water could be cooled by 5 degrees Celsius by the formation of 1 mol of diamond, we need to consider the heat capacity of water.

The heat capacity of water is approximately 4.18 J/(g·°C) or 4.18 J/g°C.

To convert the heat absorbed in kilojoules to joules:
-158.33 kJ * 1000 J/kJ = -158330 J

To determine the grams of water that can be cooled:
grams of water = heat absorbed (J) / (heat capacity (J/g·°C) * temperature change (°C))
grams of water = -158330 J / (4.18 J/g°C * 5°C) ≈ -756.22 g

Keep in mind that the negative sign indicates an exothermic process, and it means that the water will be heated, not cooled, by the formation of 1 mol of diamond.

Therefore, approximately 756.22 grams of water could be heated by 5 degrees Celsius by the formation of 1 mol of diamond.