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 a Hess's Law approach.
Step 1: Determine the heat of formation of CO2(g) using the given equations:
C(graphite) + O2 ---> CO2(g) ΔH = -393.5 kJ/mol
C(diamond) + O2(g) ---> CO2(g) ΔH = -395.4 kJ/mol
Step 2: Subtract the heat of formation of CO2 from the equation for graphite to diamond conversion to get the heat absorbed in the reaction:
C(graphite) ---> C(diamond) ΔH = -395.4 kJ/mol - (-393.5 kJ/mol)
C(graphite) ---> C(diamond) ΔH = -2.0 kJ/mol
So, the heat absorbed in the change of one mole of graphite to one mole of diamond is -2.0 kJ/mol.
Next, we can calculate the energy absorbed in the manufacture of one kilogram of diamond. To do this, we need to know the molar mass of diamond.
The molar mass of carbon (C) is approximately 12.01 g/mol. Since diamond is pure carbon, its molar mass is also 12.01 g/mol.
So, the number of moles in one kilogram of diamond is:
Number of moles = mass (in grams) / molar mass
Number of moles = 1000 g / 12.01 g/mol ≈ 83.26 mol
Now, we can calculate the energy absorbed in the manufacture of one kilogram of diamond:
Energy absorbed = Number of moles × ΔH
Energy absorbed = 83.26 mol × -2.0 kJ/mol
Energy absorbed = -166.52 kJ
Therefore, the energy absorbed in the manufacture of one kilogram of diamond is approximately -166.52 kJ.
Lastly, if the reaction were to take place in water, we need to calculate the amount of water that could be cooled by 5 degrees Celsius with the formation of 1 mole of diamond.
The specific heat capacity of water is approximately 4.18 J/(g·°C).
Let's assume the mass of water is 'm' grams.
The heat absorbed in cooling water by 5 degrees Celsius is given by:
Heat absorbed = mass × specific heat capacity × change in temperature
Heat absorbed = m g × 4.18 J/(g·°C) × 5°C
To find the mass of water, we need to convert kJ to J (1 kJ = 1000 J) and use the energy absorbed calculated earlier:
Heat absorbed = -166.52 kJ × 1000 J/kJ
Heat absorbed = -166520 J
Now we can solve for the mass of water:
-166520 J = m g × 4.18 J/(g·°C) × 5°C
m g = -166520 J / (4.18 J/(g·°C) × 5°C)
m g ≈ 7935.41 g
Therefore, approximately 7935.41 grams (or 7.94 kg) of water could be cooled by 5 degrees Celsius by the formation of 1 mole of diamond.
To determine the amount of heat absorbed in the change of one mole of graphite to one mole of diamond, we can use Hess's Law.
Hess's Law states that the overall enthalpy change of a reaction is independent of the pathway between the initial and final states and depends solely on the initial and final states of the reaction.
First, we need to find a pathway that cancels out the intermediates and gives us the desired reaction: C(graphite) ---> C(diamond). By combining the two given reactions, we can cancel out the CO2(g) since it appears on both sides of the equations:
C(graphite) + O2 ---> CO2(g) (Equation 1)
C(diamond) + O2 ---> CO2(g) (Equation 2)
When we add Equation 1 and Equation 2 together, we get:
C(graphite) + O2 + C(diamond) + O2 ---> CO2(g) + CO2(g)
This simplifies to:
C(graphite) + C(diamond) + 2O2 ---> 2CO2(g)
Now we need to determine the enthalpy change for this combined reaction. The enthalpy change for the combined reaction is the sum of the enthalpy changes of the individual reactions:
Delta H(combined) = delta H(Equation 1) + delta H(Equation 2)
Delta H(combined) = -393.5 kJ/mol + (-395.4 kJ/mol)
Delta H(combined) = -788.9 kJ/mol
Therefore, the amount of heat absorbed in the change of one mole of graphite to one mole of diamond is -788.9 kJ/mol.
To determine the amount of energy absorbed in the manufacture of one kilogram of diamond, we need to convert the given enthalpy change from kJ/mol to kJ/g.
Since the molar mass of carbon is approximately 12 g/mol, we can calculate the energy absorbed for one gram of carbon:
Energy absorbed per gram of carbon = (-788.9 kJ/mol) / (12 g/mol)
Energy absorbed per gram of carbon = -65.74 kJ/g
Now, we convert this energy absorbed per gram of carbon to energy absorbed for one kilogram of diamond:
Energy absorbed for one kilogram of diamond = (-65.74 kJ/g) * (1000 g/kg)
Energy absorbed for one kilogram of diamond = -65,740 kJ/kg
Therefore, the amount of energy absorbed in the manufacture of one kilogram of diamond is approximately -65,740 kJ/kg.
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 use the specific heat capacity of water and the amount of energy absorbed in the formation of 1 mol of diamond.
The specific heat capacity of water is approximately 4.18 J/g°C.
To convert the energy absorbed from kJ/mol to J/mol, we multiply by 1000:
Energy absorbed per 1 mol of diamond = (-788.9 kJ/mol) * (1000 J/kJ)
Energy absorbed per 1 mol of diamond = -788,900 J/mol
Now, we can calculate the amount of grams of water that can be cooled by this amount of energy:
Energy absorbed per 1 mol of diamond = specific heat capacity of water * mass of water * temperature change
Rearranging the equation for mass of water:
Mass of water = (Energy absorbed per 1 mol of diamond) / (specific heat capacity of water * temperature change)
Mass of water = (-788,900 J/mol) / (4.18 J/g°C * 5°C)
Mass of water = -37,775.84 g
Since mass cannot be negative, we take the absolute value:
Mass of water = 37,775.84 g
Therefore, approximately 37,775.84 grams of water could be cooled by 5 degrees Celsius by the formation of 1 mol of diamond.