To begin the experiment, 1.11g of methane ch4 is burned in a bomb
calorimeter containing 1000 grams of water. the initial temperature of water is 24.85oc. the specific heat of water is 4.184 j/g oc. the heat capacity of the calorimeter is 695 j/ oc . after the reaction the final temperature of the water is 35.65oc.

2. calculate the change in temperature, δt

3. using the formula qwater = m • c • δt ,calculate the heat absorbed by the water.

4. using the formula qcal = ccal • δt ,calculate the heat absorbed by the calorimeter.

5. the total heat absorbed by the water and the calorimeter can be calculated
by adding the heat calculated in steps 3 and 4. the amount of heat released
by the reaction is equal to the amount of heat absorbed with the negative
sign as this is an exothermic reaction. using the formula ∆h = -(qcal qwater + ) ,
calculate the total heat of combustion.

6. evaluate the information contained in this calculation and complete the
following sentence:
this calculation shows that burning grams of methane [takes
in/gives off] energy.

7. the molar mass of methane is 16.04 g/mol. calculate the number of moles
of methane burned in the experiment.

8. what is the experimental molar heat of combustion?

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.

10.give the formula theoretical value - experimental value % error = ×100
theoretical value , calculate
the percent error for the experiment

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 %

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

2. ΔT = T₂ - T₁  = 35.65 °C - 24.85 °C = 10.80 °C

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

4. To calculate the heat absorbed by the calorimeter,

one can use the formula

Where,

C = The heat capacity of Calorimeter = 695 J/ °C

So, Using the above formula

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.

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.

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

8. The experimental molar heat of combustion can be calculated from the following:

∴ 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.

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.

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 %

am i too late?

explanation: