When 1.23g of ethanol (delta c Hm = -1368kj/mol at 25 C ) is burned in an adiabatic flame calorimeter, the temperature increases by 8.6 C. Under the same experimental conditions, when 1.14g of 1-propanol is burned, the temperature increases by 9.0 C. What is delta c Hm for 1-propanol? Molar masses of ethanol and 1-propanol are 46.1 and 60.1g/mol, respectively.
figure the moles of each:
ethanon-1.23/46.1
prpanol-1.14/60.1
now the solution.
the temp rise is due to heat, which is Hm*molesburned
8.6/9.0=(-1366*molesethanol)/Hm*moleprpanol
calculate Hmpropaanol.
Hmprpanol=-1366*9.0/8.6*molesethanol/molespropanol
To determine the delta c Hm (change in enthalpy of combustion) for 1-propanol, we can use the heat capacity of the calorimeter and the temperature change.
Let's calculate the amount of heat released for each substance during the combustion using the equation:
q = m × c × ΔT
where:
q is the heat released (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g·°C), and
ΔT is the change in temperature (in °C).
For ethanol:
m = 1.23g
c = ? (ethanol specific heat capacity, which we can assume is similar to water, 4.18 J/g·°C)
ΔT = 8.6°C
Using the equation for ethanol:
q = 1.23g × 4.18 J/g·°C × 8.6°C = 44.6788 J
Now, let's calculate the amount of heat released for 1-propanol:
m = 1.14g
c = ? (1-propanol specific heat capacity)
ΔT = 9.0°C
Using the equation for 1-propanol:
q = 1.14g × c × 9.0°C
Since the calorimeter is considered adiabatic, the heat released during combustion is absorbed entirely by the calorimeter, and there is no heat loss to the surroundings:
q_ethanol = q_propanol
Therefore,
1.23g × 4.18 J/g·°C × 8.6°C = 1.14g × c × 9.0°C
Rearranging the equation:
c = (1.23g × 4.18 J/g·°C × 8.6°C) / (1.14g × 9.0°C)
Simplifying the equation:
c = (44.6788 J) / (1.14g × 9.0°C)
Calculating the value:
c = 43.7798 J/g·°C
Now that we have the specific heat capacity of 1-propanol, we can calculate the delta c Hm using the equation:
delta c Hm = q / n
where:
delta c Hm is the change in enthalpy of combustion per mole of substance (in kJ/mol),
q is the heat released (in J), and
n is the number of moles of the substance.
Let's calculate the delta c Hm for 1-propanol:
n = m / M
where:
m is the mass of the substance (in grams), and
M is the molar mass of the substance (in g/mol).
For 1-propanol:
m = 1.14g
M = 60.1g/mol
n = 1.14g / 60.1g/mol = 0.0189684 mol
Now we can calculate the delta c Hm for 1-propanol:
delta c Hm = 43.7798 J/g·°C / 0.0189684 mol
Calculating the value:
delta c Hm = -2,307.536 kJ/mol
Therefore, the delta c Hm for 1-propanol is approximately -2,307.536 kJ/mol.