Combustion of ethylene glycol leads to a change in internal energy (\DeltaU) of –1189 kJ at 298 K. What is the corresponding change in enthalpy in kJ\DeltaH? HOCH_2CH_2OH(l) + 5/2 O_2(g) → 2 CO_2(g) + 3 H_2O(l) \DeltaU = –1189 kJ

-1190.24 kJ

Explanation:

The enthalpy change in a chemical reaction that produces or consumes  gases is given by the expression:

ΔH = ΔU + Δngas RT

where  Δn gas is the change of moles of gas, R is the gas constant,and T is temperature.

Now from the given   balanced chemical reaction, the change in number of mol gas is equal to:

Δn gas = mole gas products - mole gas reactants =  2 - 5/2 = -1/2 mol

Sionce we know ΔU and the temperature (298 K), we are in position to calculate the change in enthalpy.

ΔH = -1189 x 10³ J + (-0.5 mol ) 8.314 J/Kmol x 298 K

ΔH = -1.190 x 10⁶ J = -1.190 x 10⁶ J x 1 kJ/1000 J = -1.190 x 10³ J

hypostyle hall;