A1.800 g sample of octane, c8h18, is burned a calorimeter whose total heat capacity is 12.66 kj/°c. if the temperature of the calorimeter increased from 22.36 °c to 28.78 °c, then what is the δh for the combustion of one mole of octane? do not add the unit in the answer.

Explanation:-

As we know heat released will be same as heat absorbed by the calorimeter

= Heat gained by bomb calorimeter

=Heat capacity of bomb calorimeter=12.66 kJ/°C

Change in temperature = ΔT=  

Let the heat released during reaction be q.

q = Heat released = 81.3 kJ

1.800 g of octane releases = 81.3 kJ of heat

Moles of octane =

0.01579 moles of octane releases 81.3 kJ of heat

1 mole of octane releases = of heat

Thus  for the combustion of one mole of octane is

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

answer7,570,82356 cubic centimeters;