**When 0.187 g of benzene,C6H6, is burned in a bomb calorimeter, the surrounding water bath rises in temperature by 7.48 C. **
Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate combustion energies (Delta E) for benzene in both kilojoules per gram and kilojoules per mole.
Well, well, well, looks like benzene really knows how to heat things up! Let's do some calculations, shall we?
First, let's find the heat absorbed by the water bath. We can use the formula:
q = mcΔT
where q is the heat absorbed, m is the mass of the water (250.0 g), c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature (7.48°C).
Plugging in the values, we get:
q = (250.0 g) * (4.18 J/g°C) * (7.48°C)
q = 7,820 J
Now, since we want to find the combustion energy (ΔE) for benzene in both kilojoules per gram and kilojoules per mole, we need to convert jiggly-joules to kooky-kilojoules.
To get the combustion energy in kilojoules per gram, we divide the heat absorbed by the mass of benzene burned (0.187 g):
ΔE (kJ/g) = (7820 J) / (0.187 g)
ΔE (kJ/g) ≈ 41,712 kJ/g
Now, to find the combustion energy in kilojoules per mole, we need to know the molar mass of benzene (C6H6). Lucky for us, it's approximately 78 g/mol.
To find the number of moles of benzene burned, we divide the mass of benzene burned by the molar mass:
n = (0.187 g) / (78 g/mol)
n ≈ 0.00239 mol
Finally, we can find the combustion energy in kilojoules per mole by dividing the heat absorbed by the number of moles:
ΔE (kJ/mol) = (7820 J) / (0.00239 mol)
ΔE (kJ/mol) ≈ 3,271,963 kJ/mol
So, in conclusion, the combustion energy for benzene is approximately 41,712 kJ/g and 3,271,963 kJ/mol. Don't you just love all those kilojoules flying around?
To calculate the combustion energy (ΔE) for benzene in kilojoules per gram and kilojoules per mole, we need to use the given mass of benzene and the temperature change of the water bath.
First, let's calculate the energy transferred to the water bath:
Energy transferred = mass of water x specific heat capacity x temperature change
Given:
Mass of water (m) = 250.0 g
Specific heat capacity of water (c) = 4.18 J/g°C
Temperature change (ΔT) = 7.48°C
Energy transferred = 250.0 g × 4.18 J/g°C × 7.48°C
Energy transferred = 7847 J
Now, let's convert the mass of benzene to moles using its molar mass:
Molar mass of benzene (C6H6) = (6 x atomic mass of carbon) + (6 x atomic mass of hydrogen)
Atomic mass of carbon = 12.01 g/mol
Atomic mass of hydrogen = 1.01 g/mol
Molar mass of benzene = (6 x 12.01 g/mol) + (6 x 1.01 g/mol)
Molar mass of benzene = 78.11 g/mol
Moles of benzene = Mass of benzene / Molar mass of benzene
Moles of benzene = 0.187 g / 78.11 g/mol
Moles of benzene = 0.00239 mol
Now, let's calculate the combustion energy (ΔE) in kilojoules:
ΔE = Energy transferred / Moles of benzene
ΔE = 7847 J / 0.00239 mol
ΔE = 3282610.04 J/mol
To convert to kilojoules per mole:
ΔE = 3282610.04 J/mol / 1000 J/1 kJ
ΔE = 3282.61 kJ/mol
Now, let's calculate the combustion energy (ΔE) in kilojoules per gram:
ΔE = Energy transferred / Mass of benzene
ΔE = 7847 J / 0.187 g
ΔE = 41946.52 J/g
To convert to kilojoules per gram:
ΔE = 41946.52 J/g / 1000 J/1 kJ
ΔE = 41.95 kJ/g
Therefore, the combustion energy for benzene is approximately 3282.61 kJ/mol and 41.95 kJ/g.
To calculate the combustion energy (ΔE) for benzene in kilojoules per gram and kilojoules per mole, we need to use the given information about the mass of benzene and the change in temperature of the surrounding water bath.
1. Calculate the heat absorbed by the water bath (q):
Using the formula q = mcΔT, where q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature, we can calculate the heat absorbed by the water bath.
Given:
- Mass of water (m) = 250.0 g
- Change in temperature (ΔT) = 7.48 °C
- Specific heat capacity of water (c) = 4.184 J/g°C (This value is given, but we will convert it to kJ/g°C for the final answer calculation)
Convert ΔT to Kelvin:
ΔT(K) = ΔT(°C) + 273.15
ΔT(K) = 7.48 + 273.15 = 280.63 K
Calculate q:
q = (250.0 g) x (4.184 J/g°C) x (280.63 K)
q = 293379.6 J
2. Convert the heat absorbed to kilojoules (kJ):
1 kJ = 1000 J
q(kJ) = q(J) / 1000
q(kJ) = 293379.6 J / 1000
q(kJ) = 293.4 kJ
3. Calculate the combustion energy (ΔE) for benzene:
Given:
- Mass of benzene = 0.187 g
4. Calculate the combustion energy per gram:
ΔE(g) = q(kJ) / mass of benzene (g)
ΔE(g) = 293.4 kJ / 0.187 g
ΔE(g) ≈ 1567.38 kJ/g
5. Calculate the molar mass of benzene (C6H6):
Molar mass (C6H6) = (6 x molar mass of carbon) + (6 x molar mass of hydrogen)
Molar mass (C6H6) = (6 x 12.011 g/mol) + (6 x 1.008 g/mol)
Molar mass (C6H6) ≈ 78.114 g/mol
6. Calculate the combustion energy per mole:
ΔE(mol) = ΔE(g) / molar mass (C6H6)
ΔE(mol) ≈ 1567.38 kJ/g / 78.114 g/mol
ΔE(mol) ≈ 20.06 kJ/mol
Therefore, the combustion energy (ΔE) for benzene is approximately 1567.38 kJ per gram and 20.06 kJ per mole.
qv = mass H2O x specific heat x delta T = delta E.
This answer will be in J/g. I assume you know how to convert to kJ/g and kJ/mole.