Substitute natural gas (SNG) is a gaseous mixture containing CH4(g) that can be used as a fuel. One reaction for the production of SNG is
4CO(g) + 8H2(g) → 3CH4(g) + CO2(g) + 2H2O(l) ΔHo = ?
Use appropriate data from the following list to determine ΔHo for this SNG reaction.
C(graphite) + 1/2O2(g) → CO(g) ΔHo = -110.5 kJ
CO(g) + 1/2O2(g) → CO2(g) ΔHo = -283.0 kJ
H2(g) + 1/2O2(g) → H2O(l) ΔHo = -285.8 kJ
C(graphite) + 2H2(g) → CH4(g) ΔHo = -74.81 kJ
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔHo = -890.3 kJ
Answer:
ΔH° of the reaction is -747.54kJ
Explanation:
Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.
Using the reactions:
(1) C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ
(2) CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ
(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ
(4) C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ
(5) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ
The sum of 4×(4) + (5) gives:
4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)
ΔH° = -74.81 kJ ×4 - 890.3 kJ = -1189.54kJ
Now, this reaction - 4×(1) gives:
4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)
ΔH° = -1189.54kJ - 4×-110.5 = -747.54kJ
Thus ΔH° of the reaction is -747.54kJ
ΔH° of the reaction is for the production of SNG -747.54kJ
Hess law and enthalpyAccording to Hess’ law of constant summation, the standard reaction enthalpy is independent of the pathway or number of the reaction steps taken for a multistep reaction, rather it is the sum of standard enthalpies of intermediate reactions involved at the same temperature.
Based on Hess law, it is possible to find the ΔH of a reaction using ΔH of half reactions.
From the given reactions:
C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJCO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJH₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJC(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJCH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJThe sum of Reaction 4 × 4 + Reaction 5 - Reaction 1 × 4 gives the reaction below:
4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)
ΔH° = {-74.81 kJ × 4} - 890.3 kJ {- 4 ×-110.5}
ΔH° = -747.54kJ
Therefore, ΔH° of the reaction is for the production of SNG -747.54kJ
Learn more about enthalpy and Hess' law at: https://brainly.com/question/9328637
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Answer:
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