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Amiraneli [1.4K]
4 years ago
5

Click the DeltaH is an Extensive Property button within the activity, and analyze the relationship between the two reactions tha

t are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the change in enthalpy for a reaction, ΔH, is an extensive property. Using this property, calculate the change in enthalpy for Reaction 2. Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ Reaction 2: 6C3H8(g)+30O2(g)→18CO2(g)+24H2O(g), ΔH2=? Express your answer to four significant figures and include the appropriate units.
Chemistry
1 answer:
DaniilM [7]4 years ago
6 0

Answer:

∴ΔH₂ = - 12,258 KJ

Explanation:

Enthalpy:

Enthalpy is a property of a thermodynamic system. Enthalpy of a system is equal to the sum of internal energy of the system and presser times volume of the system.

    The heat absorbes or releases in a closed system is the change of enthalpy of the system.

Given reactions are:

Reaction 1: C₃H₈(g)+5O₂(g)→ 3CO₂(g)+4H₂O,                 ΔH₁= - 2043 KJ

Reaction 2: 6C₃H₈(g)+30 O₂(g)→ 18 CO₂(g)+24 H₂O,     ΔH₂=?

Take a look at reaction 1 and reaction 2, the only difference is that 1 molecule of C₃H₈ is combusted in reaction 1 and 6 molecules of C₃H₈  is  combusted in reaction 2.

We can think the reaction 2 as occurring 6 different container and each containers contains 1 molecule of C₃H₈. The enthalpy is an extensive property. Total enthapy of the 6 containers is = 6×(-2043 KJ)

                                                                            = - 12,258 KJ

∴ΔH₂ = - 12,258 KJ

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<u>Step 1:</u> Data given

Mass of ammonia = 122.5 kg

Mass of carbon dioxide = 211.4 kg

Mass of urea produced = 170.3 kg

Molar mass of ammnoia = 17.031 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Moalr mass of urea = 60.06 g/mol

<u>Step 2:</u> The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

<u>Step 3:</u> Calculate moles of NH3

Number of moles = mass / Molar mass

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Moles NH3 = 7192.77 moles

<u>Step 4:</u> Calculate moles of CO2

Moles CO2 = 211400 / 44.01 g/mol

Moles CO2 = 4803.45 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

NH3 is the limiting reactant. It will completely be consumed (7192.77 moles).

CO2 is in excess. There will be consumed 7192.77/2 = 3596.4 moles

There will remain 4803.45 - 3596.4 = 1207.05 moles of CO2

<u>Step 6:</u> Calculate moles of urea produced:

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

For 7192.77 moles of NH3, we have 3596.4 moles of urea produced

<u>Step 7: </u>Calculate mass of urea

Mass urea = moles urea * molar mass urea

Mass urea = 3596.4 moles * 60.06 g/mol

Mass urea = 216000 grams = 216 kg = theoretical yield

<u>Step 8</u>: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (170.3 / 216) *100% = 78.8%

The % for this reaction is 78.8 %

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