Chemistry: 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

To begin the experiment, 1.11g of methane ch4

Ответ:

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15.06.2020

Chemistry

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 = $Q=mc\Delta T=1000 g\times 4.184 J/g^oC\times 10.80^oC=45,187.2 J$

Heat absorbed by the calorimeter = $Q_{cal}=c_{cal}\Delta T=695 J/^oC\times 10.80^oC=7,506 J$

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 = $\frac{1.11 g}{16.04 g/mol}=0.0692 mol$

Experimental molar heat of combustion = $\frac{52,693.2 J}{0.0692 mol}=-761,462.42 j/mol=-761.462 kJ/mol$

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:

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