A sample of ethanol, C2H5OH, weighing 2.84 g was burned in an excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rose from 25.00°C to 33.73°C. If the heat capacity of the calorimeter and contents was 9.63 kJ/°C, what is the value of q for burning 1.28 mol of ethanol at constant volume and 25.00°C?
The reaction is shown below.
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)
q for calorimeter is heat cap x delta T.
qcal = 9.63 kJ/C x (33.73-25.00) = 84.1 kJ and that is for 2.84 g ethanol. I would convert 2.84 g ethanol to moles.
2.84/46 = 0.0617 moles; therefore, the heat of burning is 84.1 kJ/0.0617 = 1,362 kJ/mol. Now all you need to do is determine the amount of heat from burning 1.28 mol since you have the heat for 1 mole. Check my work above and round everything to the appropriate number of significant figures. Also, I estimated the molar mass of C2H5OH. It may not be exactly 46.0.
Well, it seems like you want to know the value of q for burning 1.28 mol of ethanol. Let me calculate that for you, but first, I have to ask... Did the ethanol at least go out in a blaze of glory?
Now, let's work on this seriously clownish question. We can find the heat released by the reaction by using the heat capacity of the calorimeter and the change in temperature.
The change in temperature is 33.73°C - 25.00°C = 8.73°C.
To find the heat released, we can use the formula:
q = C * ΔT
where q is the heat released, C is the heat capacity of the calorimeter and contents, and ΔT is the change in temperature.
Plugging in the values, we get:
q = 9.63 kJ/°C * 8.73°C
Calculating this, we find:
q = 84.15 kJ
Now, since we burned 1.28 mol of ethanol, we need to calculate the heat released per mole. We can do this by dividing the total heat released by the number of moles:
q_per_mole = q / n
where q_per_mole is the heat released per mole, q is the total heat released, and n is the number of moles of ethanol.
Plugging in the values, we get:
q_per_mole = 84.15 kJ / 1.28 mol
Calculating this, we find:
q_per_mole ≈ 65.86 kJ/mol
So, the value of q for burning 1.28 mol of ethanol at constant volume and 25.00°C is approximately 65.86 kJ/mol. Quite a fiery performance, if you ask me!
To find the value of q for burning 1.28 mol of ethanol at constant volume and 25.00°C, we can use the equation:
q = ΔT * C
where q is the heat transferred, ΔT is the change in temperature, and C is the heat capacity of the calorimeter and contents.
First, let's calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 33.73°C - 25.00°C
ΔT = 8.73°C
Next, let's convert the heat capacity from kJ/°C to J/°C:
C = 9.63 kJ/°C * 1000 J/1 kJ
C = 9630 J/°C
Now, substitute the values into the equation:
q = ΔT * C
q = 8.73°C * 9630 J/°C
q = 84703.1 J
Since we know that the molar mass of ethanol (C2H5OH) is 46.07 g/mol, we can convert the mass of ethanol burned from grams to moles:
moles of ethanol = 2.84 g / 46.07 g/mol
moles of ethanol = 0.061731 mol
Finally, we can calculate the value of q for burning 1.28 mol of ethanol:
q for 1.28 mol of ethanol = q * (1.28 mol / 0.061731 mol)
q for 1.28 mol of ethanol = 84703.1 J * 20.751816
q for 1.28 mol of ethanol = 1,759,362.84 J
Therefore, the value of q for burning 1.28 mol of ethanol at constant volume and 25.00°C is approximately 1,759,362.84 J.
To calculate the value of q for burning 1.28 mol of ethanol at constant volume and 25.00°C, we can use the equation:
q = m × ΔT × C
where:
q is the heat released or absorbed during the reaction,
m is the number of moles of the substance being burned (in this case, ethanol),
ΔT is the change in temperature of the calorimeter,
C is the heat capacity of the calorimeter and contents.
First, we need to calculate the energy released (q) when 1 mol of ethanol is burned. The balanced equation shows that for every 1 mol of ethanol, 2 mol of CO2 is produced. The enthalpy change (ΔH) for the combustion of ethanol to produce CO2 is -1368 kJ/mol.
Since we want the value of q for 1.28 mol of ethanol, we multiply -1368 kJ/mol by 1.28 mol to get the energy released.
q = -1368 kJ/mol × 1.28 mol = -1749.12 kJ
Next, we need to calculate the heat capacity (C) of the calorimeter and contents in J/°C. We are given that C = 9.63 kJ/°C. We need to convert kJ to J by multiplying by 1000:
C = 9.63 kJ/°C × 1000 J/1 kJ = 9630 J/°C
Now we can use the equation to find the value of q for burning 1.28 mol of ethanol:
q = m × ΔT × C
We are given that the temperature change (ΔT) is 33.73°C - 25.00°C = 8.73°C.
q = 1.28 mol × 8.73°C × 9630 J/°C = 108,800 J
Finally, we can convert the value of q to kilojoules by dividing by 1000:
q = 108,800 J / 1000 = 108.8 kJ
So, the value of q for burning 1.28 mol of ethanol at constant volume and 25.00°C is 108.8 kJ.