Am I correct?
From data below, calculate the total heat (J) needed to convert 0.539 mol gaseous ethanol at 300°C and 1 atm to liquid ethanol at 25.0°C and 1 atm.
Boiling point at 1 atm 78.5°C
c ethanol gas 1.43 J/g·°C
c ethanol liquid 2.45 J/g·°C
ÄH°vap 40.5 kJ/mol
help please Answer in J
my work:
.539 x 46= 24.79
24.79 x 1.43 x -221.50= -7852 J
.539 x -40.5= -21,829 J
24.79 x 2.45 x -53.5 = 32930 J
then -7852 + -21, 829 + -3249= -32930 J
539 x 46= 24.79
24.79 x 1.43 x -221.50= -7852 J
.539 x -40.5= -21,829 J
24.79 x 2.45 x -53.5 = 32930 J
I would have rounded step 2 and step 3 differently but other than that I don't have any problems with steps 1, 2, or 3.
For step 4 I think you made a typo here and put the answer of 32930 J for total instead of -3249 J for this tep. I think that answer should be -3249 J.
I'm a little confused about what the question actually wants. If the question want's total heat (liberated) when ethanol goes from 300 to 25 then your 32930 J is correct. If the question wants total heat from 300 to 25 then -32930 J is right but I notice the problem actually says "how much is NEEDED in ....". Actually there is no heat needed since each of these steps is exothermic each step gives off heat and none is needed. ;-)
To calculate the total heat needed to convert gaseous ethanol to liquid ethanol, we need to consider the following steps:
Step 1: Calculate the heat needed to cool the gaseous ethanol from 300°C to its boiling point at 78.5°C.
Heat = mass × specific heat capacity × change in temperature
Heat = (mass of ethanol × specific heat capacity of ethanol gas × change in temperature)
= (0.539 mol × 46 g/mol) × (1.43 J/g·°C) × (300°C - 78.5°C)
= 7852 J (remember to use positive values for temperature differences)
Step 2: Calculate the heat needed for the phase change from gaseous ethanol at its boiling point to liquid ethanol at the same temperature.
Heat = (mol of ethanol × heat of vaporization)
= (0.539 mol × 40.5 kJ/mol × 1000 J/kJ)
= 21,829 J
Step 3: Calculate the heat needed to cool the liquid ethanol from 78.5°C to 25.0°C.
Heat = (mass of ethanol × specific heat capacity of ethanol liquid × change in temperature)
= (0.539 mol × 46 g/mol) × (2.45 J/g·°C) × (78.5°C - 25.0°C)
= 32,930 J
Step 4: Add up the results from the previous steps to get the total heat needed.
Total heat = (7852 J) + (21,829 J) + (32,930 J)
= 62,611 J
Therefore, the total heat needed to convert gaseous ethanol at 300°C and 1 atm to liquid ethanol at 25.0°C and 1 atm is 62,611 J.
To calculate the total heat needed to convert gaseous ethanol to liquid ethanol, you need to consider two processes: heating the gaseous ethanol to its boiling point and then condensing it to liquid at a lower temperature.
Firstly, let's calculate the heat required to raise the temperature of 0.539 moles of gaseous ethanol from 300°C to its boiling point at 78.5°C.
The molar heat capacity of the gas ethanol (c_ethanol_gas) is given as 1.43 J/g·°C. To convert this to the molar heat capacity, we need to multiply it by the molar mass of ethanol (46 g/mol):
1.43 J/g·°C * 46 g/mol = 65.78 J/mol·°C
Next, calculate the temperature difference (ΔT_gas) between the initial temperature (300°C) and the boiling point (78.5°C):
ΔT_gas = 78.5°C - 300°C = -221.5°C
Now, you can calculate the heat required to raise the temperature of gaseous ethanol:
q1 = n_ethanol_gas * c_ethanol_gas * ΔT_gas
= 0.539 mol * 65.78 J/mol·°C * -221.5°C
= -7852 J
Now, let's calculate the heat required to condense the gaseous ethanol at its boiling point (78.5°C) to liquid ethanol at 25.0°C using the enthalpy of vaporization (ΔH_vap) given as 40.5 kJ/mol.
Convert the enthalpy of vaporization to joules per mole:
ΔH_vap = 40.5 kJ/mol * 1000 J/kJ
= 40,500 J/mol
Next, calculate the heat required to condense the gaseous ethanol:
q2 = n_ethanol_gas * ΔH_vap
= 0.539 mol * 40,500 J/mol
= -21,829 J
Finally, calculate the heat required to cool the gaseous ethanol to 25.0°C using the molar heat capacity of liquid ethanol (c_ethanol_liquid) given as 2.45 J/g·°C:
The molar heat capacity is already in joules per mole per degree Celsius, so no conversion is needed.
Calculate the temperature difference (ΔT_liquid) between the initial boiling point (78.5°C) and the final temperature (25.0°C):
ΔT_liquid = 78.5°C - 25.0°C
= -53.5°C
Now, calculate the heat required to cool the gaseous ethanol:
q3 = n_ethanol_gas * c_ethanol_liquid * ΔT_liquid
= 0.539 mol * 2.45 J/mol·°C * -53.5°C
= -32930 J
To find the total heat needed, sum up the three heat values calculated:
Total heat = q1 + q2 + q3
= -7852 J + -21,829 J + -32930 J
= -68,611 J
Therefore, the total heat needed to convert 0.539 moles of gaseous ethanol at 300°C and 1 atm to liquid ethanol at 25.0°C and 1 atm is -68611 J.