Chemistry, 15.11.2019 20:31 johnnybones03

The enthalpy of formation of liquid ethanol (c2h5oh) is −277.6 kj/mol. what is the equation that represents the formation of liquid ethanol? a. 2 c(s) + 6 h(g) + o(g) → c2h5oh(l) b. 2 c(s) + 3 h2(g) + ½ o2(g) → c2h5oh(l) c. 2co2(g) + 3h2o(g) → c2h5oh(l) + 3 o2(g) d. 4 c(s) + 6 h2(g) + o2(g) → 2 c2h5oh(l)

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Answer from: milamdaniela

B. 2 C(s) + 3 H₂(g) + ½ O₂(g) → C₂H₅OH(l)

Explanation:

The enthalpy of formation refers to the formation of 1 mole of a substance from the simple substances in their most stable states. The thermochemical equation that represents this reaction is:

2 C(s) + 3 H₂(g) + ½ O₂(g) → C₂H₅OH(l) ΔH°f = -277.6kJ/mol

Answer from: zaida00

b. $2 C(s) + 3 H_2(g) + \frac{1}{2} O_2(g) \rightarrow C_2H_5OH(l)$

Explanation:

Hello,

In this case, formation reactions are said to be undergone when only the elements constituting the involved compounds, react in order to yield it in a combination chemical reaction. In such a way, for ethanol, carbon, hydrogen and oxygen must react to form it. In addition, it is mandatory to remember that hydrogen and oxygen only exist diatomic as pure gaseous elements. Moreover, the reaction must be properly balanced therefore, ethanol's formation reaction turns out:

$2 C(s) + 3 H_2(g) + \frac{1}{2} O_2(g) \rightarrow C_2H_5OH(l)$

Which have an enthalpy of formation of -277.6 kJ/mol (exothermic reaction).

Best regards.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

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The enthalpy of formation of liquid ethanol (c2h5oh) is −277.6 kj/mol. what is the equation that rep...

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