Question

17. Determine the enthalpy change for the oxidation of ethanol to acetic acid, CH3CH2OH(() + O2(g) → CH3COOH() + H2O(1) given the thermochemical equations below. 1) 2 CH2CH2OH(1) + O2(g) + 2 CH CHO(l) + 2 H20(1) AH = -400.8 kJ/mol-rxn 2) 2 CHỊCHO(0)+0g)+2 CHCOOH(C). A,Hº = -584.4 kJ/mol-rxn (a) What is the compound that is not present in the target equation? (b) To cross out that compound, what operation is needed to do for equations 1) and 2)? (c) What's the A. Hº of the target equation? (Check the stoichiometric coefficient to match the original equation) 18. The decomposition of KCIO3 will produce KCl and O2 KCIO;(s) → KCl(s) + O2(g) (a) Write the balanced equation for this reaction. (b) How many moles of O2 will be produced from 5.00 g KCIO. (c) What is the volume of that O2 measured at 225 °C and 0.970 atm (R=0.08206 L.atm/mol-K)?

Answer 1

Question 17

Given

Target equation : CH₃CH₂OH (l) + O ₂ (g) $\rightarrow$ CH₃COOH (l) + H₂O (l)

Thermochemical equations

1) 2 CH₃CH₂OH (l) + O ₂ (g) $\rightarrow$ 2 CH₃CHO (l) + 2 H₂O (l)   $\Delta$rH ⁰ = - 400.8 k J / mol

2) 2 CH₃CHO (l) + O ₂ (g) $\rightarrow$ 2 CH₃COOH (l) $\Delta$ rH ⁰ = - 584.4 kJ / mol

a) Compound CH₃CHO is not present in the target equation.

b) To cross out the compound CH₃CHO we need to add equation 1 and equation 2

C)

Adding equation 1 and equation 2 , we get equation

2 CH₃CH₂OH (l) + O ₂ (g) $\rightarrow$ 2 CH₃CHO (l) + 2 H₂O (l)

+ 2 CH₃CHO (l) + O ₂ (g) $\rightarrow$ CH₃COOH (l)

Overall equation : 2 CH₃CH₂OH (l) + 2 O ₂ (g) $\rightarrow$ 2 CH₃COOH (l) + 2 H₂O (l)

$\Delta$rH ⁰ = ( - 400.8 k J / mol ) + ( - 584.4 kJ / mol ) = - 985.2 k J / mol

To get target equation , multiply above equation by 1/2.

$\therefore$ CH₃CH₂OH (l) + O ₂ (g) $\rightarrow$ CH₃COOH (l) + H₂O (l)  $\Delta$rH ⁰ = 1/2 ( - 985.2 k J / mol ) = - 492.6 kJ / mol

ANSWER : $\Delta$ rH ⁰ = - 492.6 kJ / mol