4.41 g of propane gas (C3H8) is injected into a bomb calorimeter and ignited with excess oxygen, according to the reaction below. The calorimeter (including the water) has a heat capacity of 97.1 kJ/°C. C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O() (a) If the temperature rose from 25.000°C to 27.282°C, what is the heat of the reaction, qrxn?

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Alleei Alleei · Answer 1 · 2019-11-21T07:50:04+01:00

Answer : The heat of the reaction is -221.6 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter

[tex]q_{rxn}=-q_{cal}[/tex]

[tex]q_{cal}=c_{cal}\times \Delta T[/tex]

where,

[tex]q_{rxn}[/tex] = heat released by the reaction = ?

[tex]q_{cal}[/tex] = heat absorbed by the calorimeter

[tex]c_{cal}[/tex] = specific heat of calorimeter = [tex]97.1kJ/^oC=97100J/^oC[/tex]

[tex]\Delta T[/tex] = change in temperature = [tex](T_{final}-T_{initial})=(27.282-25.000)=2.282^oC[/tex]

Now put all the given values in the above formula, we get:

[tex]q_{cal}=(97100J/^oC)\times (2.282^oC)[/tex]

[tex]q_{cal}=221582.2J=221.6kJ[/tex]

As, [tex]q_{rxn}=-q_{cal}[/tex]

So, [tex]q_{rxn}=-221.6kJ[/tex]

Thus, the heat of the reaction is -221.6 kJ