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### Topic: Can someone help with this question??  (Read 313 times)

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#### Rudul

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##### Can someone help with this question??
« on: February 28, 2021, 05:27:52 PM »
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).

#### Borek

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##### Re: Can someone help with this question??
« Reply #1 on: February 28, 2021, 06:08:39 PM »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Vidya

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##### Re: Can someone help with this question??
« Reply #2 on: March 01, 2021, 01:02:30 AM »

Using only this data calculate the molar enthalpy of formation for ethanol, C2H5OH(ℓ).
Do you know meaning of enthalpy of formation?
Once that is clear write equation which represents formation of 1 mole of ethanol.
Now use Hess's law on the given equations so that when you add all of them you get reaction representing the formation enthalpy of ethanol.

#### Orcio_Dojek

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##### Re: Can someone help with this question??
« Reply #3 on: March 06, 2021, 02:21:18 PM »
@Rudul - we know that:

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

(b) C(s) + O2(g) → CO2(g)     ΔH = -395.3kJ

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

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

and we are looking for:

(e) 2 C + 3 H2 + 1/2 O2 → C2H5OH

If you add (a) + 2(b) + 1,5(c) - (d) then you will get this:

CH2=CH2 + H2O + 2 C + 2 O2 + 3 H2 + 1,5 O2 + 2 CO2 + 2 H2O → C2H5OH + 2 CO2 + 3 H2O + CH2=CH2 + 3 O2

which is equal to:

(e) 2 C + 3 H2 + 1/2 O2 → C2H5OH

Now you have to calculate the result of reaction (e) (- 518 kJ / mol).
« Last Edit: March 06, 2021, 05:08:56 PM by Orcio_Dojek »