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
I thought Bob Pursley showed you how to do this.
Add equation to the reverse of equation 2. That gives you kJ/mol graphite to diamond. If you want 1 kg, then
kJ/mol from the above x (1000 g/12.01) = ? heat absorbed to make 1 kg.
For part b.
?kg from above = mass H2O x specific heat H2O x 5
Solve for grams H2O. I estimate the answer to be about 100 grams but check me out on that.
To determine the amount of heat absorbed in the change of one mole of graphite to one mole of diamond, we need to apply the Hess's law. It states that if a reaction can be expressed as the sum of two or more other reactions, then the enthalpy change of the first reaction is equal to the sum of the enthalpy changes of the other reactions.
First, let's write down the balanced equation for the conversion of graphite to diamond:
C(graphite) → C(diamond)
Since we are given the enthalpies of the reactions involving carbon and oxygen, we need to rearrange them in order to match the desired reaction. To do this, we'll multiply the second reaction by -1:
-C(diamond) + O2(g) → -CO2(g)
Now, we can sum up the two equations to cancel out the CO2:
C(graphite) + O2(g) → C(diamond)
- C(diamond) + O2(g) → -CO2(g)
By summing these two equations, we get:
C(graphite) → C(diamond) + CO2(g)
Now, let's apply the Hess's law:
ΔH = Σ ΔH(products) - Σ ΔH(reactants)
ΔH = (-395.4 kJ/mol) - (0 kJ/mol) - (-393.5 kJ/mol)
ΔH = -395.4 kJ/mol + 393.5 kJ/mol
ΔH = -1.9 kJ/mol
So, for the conversion of one mole of graphite to one mole of diamond, -1.9 kJ of heat is absorbed.
To determine the amount of energy absorbed in the manufacture of one kilogram of diamond, we need to convert the mass of diamond from grams to moles. The molar mass of diamond is approximately 12 g/mol. Therefore, one kilogram of diamond is equal to:
1 kg = 1000 g = 1000 g / 12 g/mol ≈ 83.33 moles
Now, we can calculate the total energy absorbed:
Total energy absorbed = (-1.9 kJ/mol) × (83.33 moles)
Total energy absorbed = -158.327 kJ
Therefore, the manufacture of one kilogram of diamond absorbs approximately 158.327 kJ of energy.
To calculate how many grams of water could be cooled by 5 degrees Celsius by the formation of 1 mol of diamond, we need to apply the equation:
q = mcΔT
Where:
q is the heat absorbed or released in joules,
m is the mass of water in grams,
c is the specific heat capacity of water (4.184 J/g°C),
and ΔT is the temperature change in Celsius.
We know that the heat absorbed is equal to -1.9 kJ/mol, and the molar mass of diamond is approximately 12 g/mol. Therefore, we can rearrange the equation to solve for m:
q = (-1.9 kJ/mol)
q = (-1.9 × 10^3 J/mol)
m × (4.184 J/g°C) × 5°C = (-1.9 × 10^3 J/mol)
Solving for m:
m = (-1.9 × 10^3 J/mol) / (4.184 J/g°C × 5°C)
m ≈ -91.06 g
Since mass cannot be negative, we can assume that the formation of 1 mol of diamond can cool approximately 91.06 grams of water by 5 degrees Celsius.
To determine the amount of heat absorbed in the change of one mole of graphite to one mole of diamond, we can use the principle of Hess's Law. Hess's Law states that if a reaction can be expressed as a series of multiple reactions, the enthalpy change of the overall reaction can be calculated by summing the enthalpy changes of the individual reactions.
In this case, we can see that the direct reaction between graphite and diamond is not given. However, we have two reactions involving carbon and oxygen that can be used to indirectly determine the enthalpy change of the graphite to diamond transformation.
First, let's write the balanced equation for the formation of carbon dioxide (CO2) from graphite (C(graphite)) and oxygen (O2):
C(graphite) + O2 ---> CO2 ΔH = -393.5 kJ/mol (Reaction 1)
Next, we need to write the balanced equation for the formation of carbon dioxide (CO2) from diamond (C(diamond)) and oxygen (O2):
C(diamond) + O2 ---> CO2 ΔH = -395.4 kJ/mol (Reaction 2)
Now, let's determine the enthalpy change for the transformation of graphite to diamond (C(graphite) ---> C(diamond)).
Since the formation reaction of CO2 in Reaction 2 involves the formation of diamond, we can reverse Reaction 2 and multiply it by (-1) to get the transformation of diamond to graphite:
CO2 ---> C(diamond) + O2 ΔH = +395.4 kJ/mol
The ΔH value for this reaction is positive because we are going in the reverse direction.
Now, we can sum Reaction 1 and the reversed Reaction 2 to cancel out the CO2 and O2 terms, leaving us with the transformation from graphite to diamond:
C(graphite) + O2 + CO2 ---> C(diamond) + O2 + CO2 ΔH = -393.5 kJ/mol + (+395.4 kJ/mol)
Simplifying the equation:
C(graphite) ---> C(diamond) ΔH = +1.9 kJ/mol
Therefore, the amount of heat absorbed in the change of one mole of graphite to one mole of diamond is +1.9 kJ/mol.
To determine the amount of energy absorbed in the manufacture of one kilogram of diamond, we need to know the molar mass of diamond. The molar mass of carbon is approximately 12.01 g/mol.
First, convert one kilogram of diamond to grams:
1 kilogram = 1000 grams
Next, calculate the number of moles in one kilogram of diamond:
Number of moles = mass (in grams) / molar mass
Number of moles = 1000 grams / 12.01 g/mol
Now, multiply the number of moles by the enthalpy change per mole to get the total energy absorbed:
Energy absorbed = Number of moles * ΔH
Finally, substitute the values into the equation to calculate the amount of energy absorbed in the manufacture of one kilogram of diamond.
To determine how many grams of water could be cooled by 5 degrees Celsius by the formation of 1 mole of diamond in water, we need to consider the specific heat capacity of water.
The specific heat capacity of water is approximately 4.18 J/g°C. This means that 4.18 Joules of energy are required to raise the temperature of 1 gram of water by 1 degree Celsius.
First, convert the energy absorbed in the formation of 1 mole of diamond (calculated previously) from kJ to J:
Energy absorbed = 1.9 kJ * 1000 J/1 kJ = 1900 J
Next, divide the energy absorbed by the specific heat capacity of water to calculate the number of grams of water that can be cooled by 5 degrees Celsius:
Grams of water cooled = Energy absorbed / (specific heat capacity * temperature change)
Grams of water cooled = 1900 J / (4.18 J/g°C * 5°C)
Finally, substitute the values into the equation to calculate the number of grams of water that can be cooled.