Calculate the work (w) and ΔEo, in kJ, at 298 K and 1 atm pressure, for the combustion of one mole of C4H10 (g). First write and balance the equation. The products will be CO2 (g) and H2O (g) The value of ΔHo for this reaction is -2658.3 kJ/mol

Explanation:

2C₄H₁₀ + 13O₂ = 8 CO₂ + 10H₂O

Change in number of moles Δn = 18 - 15 = + 3 moles .

ΔHo = -2658.3 kJ/mol.

ΔHo = ΔEo+ Δn RT

Δn = 3

For one mole Δn = 1.5

ΔHo = ΔEo+ W

W = Δn RT

= 1.5 x 8.31 x 298

= 3714.5 J

= 3.7 kJ /mole

ΔHo = ΔEo+ W

ΔEo =  ΔHo -  W

= -2658.3 - 3.7  kJ

= - 2662 kJ .

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

answer carbon and three oxygen atoms;