Respuesta :
Answer:
2. ∆ T = 10.8 ˚C
3. q = 45,187.2 J
4. q = 7506 J
5. q (rxn) = - 52,693.2 J
6. This calculation shows that burning 1.11 grams of methane [takes in/gives off] gives off energy.
7. The number of moles of methane burned in the experiment = 0.069 mol.
8. Experimental molar heat of combustion = - 761,462.4 J/mol = - 761.5 kJ/mol
9. This may be attributed to experimental error occurring like thermometers misreading , heat loses to surrounding (as the efficiency of any calorimeter doesn't reach 100 % ), and poor lab technique.
10. % Error = 14.4 %
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Explanation:
Firstly, from the given information
for water:
Initial temperature = T₁ = 24.85 °C
Final temperature = T₂ = 35.65 °C
So, the the change in temperature can be calculated
ΔT = Final temperature - Initial temperature
For calorimeter:
The heat capacity of the calorimeter is 695 J/ °C
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2. ΔT = T₂ - T₁ = 35.65 °C - 24.85 °C = 10.80 °C
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3. The heat absorbed by the water can be calculated from
q(water) = m • c • ΔT
where,
m =mass of water = 1000 g
C = The specific heat capacity of water = 4.184 J/g °C
So,
q(water) = m • c • ΔT = 1000 g * 4.184 J/g °C * 10.80 °C
q(water) = 45,187.2 J
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4. To calculate the heat absorbed by the calorimeter,
one can use the formula
- q(cal) = C(cal) • ΔT ,
Where,
C = The heat capacity of Calorimeter = 695 J/ °C
So, Using the above formula
- q(cal) = C(cal) • ΔT
- q(cal) = 695 J/ °C * 10.8 °C = 7,506 J.
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5. The total heat of combustion reaction can be calculated using the formula
∵ Amount of heat released by the reaction = - [Amount of heat absorbed by Calorimeter + Amount of heat absorbed by Water]
q (rxn) = -[q(cal) + q(water)
]
∴ q (rxn) = -[7,506 J + 45,187.2 J
] = - 52,693.2 J
That means that this is an exothermic reaction.
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6. By evaluating the above information one can complete the following sentence:
This calculation shows that burning 1.11 grams of methane [takes in/gives off] gives off energy.
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7. To calculate the number of moles of methane burned in the experiment
No. of moles = (mass / molar mass) = (1.11 g / 16.04 g/mol) = 0.069 mol
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8. The experimental molar heat of combustion can be calculated from the following:
- molar heat of combustion = (q (rxn) / no. ofmoles of methane)
- where q (rxn) is the total heat of combustion calculated in step 6, and no. of moles of methane was calculated in step 7.
∴ Experimental molar heat of combustion = (q (rxn) / no. ofmoles of methane)
Experimental molar heat of combustion = (-52,693.2 J / 0.069 mol)
= - 761,462.4 J/mol
= - 761.5 kJ/mol.
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9. The accepted value for the heat of combustion of methane is -890 KJ/mol. Explain why the experimental data might differ from the theoretical value.
This may be attributed to experimental error occurring like thermometers misreading , heat loses to surrounding (as the efficiency of any calorimeter doesn't reach 100 %),and poor lab technique.
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10. Calculate the percent error for the experiment.
Given the formula below:
% Error = [Theoretical value - Experimental value) /Theoretical value] ×100
one can Calculate the percent error for the experiment
where
Theoretical value = - 890 KJ/mol.
Theoretical value = - 761.5 KJ/mol.
% Error = [Theoretical value - Experimental value) /Theoretical value] ×100
% Error = [(-890) - (-761.5) /(-890)] ×100
= 14.4 %
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