Question

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.

PLEASE HELP WITH ANY OF THEM!!

Answer 1

I am working on this now but I think 1. is CH4+2O2-->CO2+2H2O I will help more when I work on the other ones

Answer 2

Step-by-step explanation:

Initial temperature of water = 24.85°C

Final temperature of water = 35.65°C

Mass of water = 1000 g

The specific heat of water ,c = 4.184 J/g °C.

The heat capacity of the calorimeter
`c_(cal)` = 695 J/ °C

Change in temperature
`\Delta T=35.65^oC-24.85^oC=10.80^oC`

Heat absorbed by the water =
`Q=mc\Delta T=1000 g* 4.184 J/g^oC* 10.80^oC=45,187.2 J`

Heat absorbed by the calorimeter =
`Q_(cal)=c_(cal)\Delta T=695 J/^oC* 10.80^oC=7,506 J`

Total heat combustion =

Heat absorbed by the water + Heat absorbed by the calorirmeter

=(-45,187.2 J)+ (-7,506 J) = -52,693.2 J

1.11 grams of methane gives off 52,693.2 J.

Number of moles methane burned in an experiment =
`(1.11 g)/(16.04 g/mol)=0.0692 mol`

Experimental molar heat of combustion =
`(52,693.2 J)/(0.0692 mol)=-761,462.42 j/mol=-761.462 kJ/mol`

Negative sign heat is released during burning of methane.

The experimental data might differ from the theoretical value because while performing an experiment there are certain factors which affects the the observations and experimental procedure which results in differ value experimental value from theoretical value.

The percentage error in the experiment:

`|\frac{\text{Theoretical value-Experimental value}}{\text{Theoretical value}}|* 100=(|(-890 kJ/mol)-(-761.462 kJ/mol)|)/(-890 kJ/mol)* 100=14.4\%`

The percent error for the experiment is 14.4 %.