Answer to Question #167743 in General Chemistry for Rudul

Question #167743

In a calorimetry experiment the following data were obtained. 

CH2CH2(g) + H2O(ℓ) → C2H5OH(ℓ)   ΔH = -288.9kJ

C(s) + O2(g) → CO2(g)   ΔH = -395.3kJ

2 H2(g) + O2(g) → 2 H2O(ℓ)   ΔH = -587.4kJ

CH2CH2(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(ℓ)   ΔH = -1441.1kJ

Using only this data calculate the molar enthalpy of formation for ethanol, C2H5OH(ℓ).


1
Expert's answer
2021-03-03T01:32:56-0500

The required equation is

"2C(s)+3H_2(g)+\\dfrac{1}{2}O_2(g)\\longrightarrow C_2H_5OH(l)\\space\\space\\space\\space\\space\\Delta H=?"

Given equations are

"CH_2CH_2(g)+H_2O(l)\\longrightarrow C_2H_5OH(l)\\space\\space\\space\\space\\space\\Delta H_1=-288.9kJ\\space(1)"

"C(s)+O_2(g)\\rightarrow CO_2(g)\\space\\space\\space\\space\\Delta H_2=-395.3kJ\\space\\space(2)"

"2H_2(g)+O_2(g)\\rightarrow2H_2O(l)\\space\\space\\space\\space\\Delta H_3=-587.4kJ\\space\\space(3)"

"CH_2CH_2(g)+3O_2(g)\\rightarrow 2CO_2(g)+2H_2O(l)\\space\\space\\space\\space\\Delta H_4=-1441.1kJ\\space\\space(4)"


Multiplying eq. (2) by 2 and eq.(3) by 3/2 and adding

"2C(s)+3H_2(g)\\rightarrow2CO_2(g)+3H_2O(l) \\space\\space\\space\\space\\space\\space\\Delta H_5=-1671.1kJ\\space\\space(5)"


Subtracting eq.(4) from eq.(5) and then adding eq.(1)

"2C(s)+3H_2(g)+\\dfrac{1}{2}O_2(g)\\rightarrow C_2H_5OH(l)\\space\\space\\space\\space\\Delta H_f=-518.9kJ\/mol"


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