Question

Consider the combustion of 5.99 mol of liquid ethanol (C2H5OH) to gaseous water and carbon dioxide. Calculate the enthalpy change for this reaction. The pertinent enthalpies of formation (in kJ/mol) are ∆Hf H2O = −241.8 ∆Hf CO2 = −393.5 ∆Hf C2H5OH = −277.7 Answer in units of kJ.

Answer 1

Enthaply change when 5.99 moles of alcohol undergone combustion is -7395.853 kilo Joules..

Step-by-step explanation:

`H_2(g)+(1)/(2)O_2(g)\rightarrow H_2O(l),\Delta H_(f,1)=-241.8 kJ/mol`..[1]

`C(s)+O_2(g)\rightarrow CO_2(g),\Delta H_(f,2)=-393.5 kJ/mol`..[2]

`2C(g)+3H_2(g)+(1)/(2)O_2(g)\rightarrow C_2H_5OH(l),\Delta H_(f,3)=-277.7 kJ/mol`..[3]

`C_2H_5OH(l)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(l),\Delta H_(comb)=?`..[4]

2 × [2] + 3 × [1] - [3] = [4]

`\Delta H_(comb)=2* \Delta H_(f,2)+3* \Delta H_(f,1)-\Delta H_(f,3)`

`=2* (-393.5 kJ/mol) +3* (-241.8 kJ/mol) - (-277.7 kJ/mol)`

`\Delta H_(comb)=-1,234.7 kJ/mol`

Enthalpy of combustion of ethanol is -1,234.7 kJ/mol.

Enthaply change when 5.99 moles of alcohol undergone combustion:

`-1,234.7 kJ/mol* 5.99 mol=-7395.853 kJ`