Chemistry - Thermochemistry Question with Solution | TestHub
Match the columns
Column I | Column II |
|---|---|
P) C(diamond) + O2(g) → CO2(g) | i) Δr H° = Δc H° |
Q) C(graphite) + O2(g) → CO2(g) | ii) Δr H° = Δf H° |
R) H2(g) → 2H(g) | iii) Δr H° = Δatomization H° |
S) CH4(g) → C(g) + 4H(g) | iv) Δr H° = Δbond H° |
(Column I = Reaction; Column II = Δr H° equals)
Options:
Answer:
Solution:
P: Diamond is not the standard state of carbon. Therefore, burning it yields only the standard enthalpy of combustion (Δ H°c) of diamond. This corresponds to option P-i.
Q: Burning graphite represents both the combustion of carbon and the standard enthalpy of formation of CO₂. This corresponds to options Q-i and Q-ii.
R: Breaking 1 mol of H₂ into individual atoms is the atomization of hydrogen and is exactly equal to the H–H bond enthalpy. This corresponds to options R-iii and R-iv.
S: The reaction CH₄(g) → C(g) + 4H(g) represents the atomization of methane. This value equals four times the average C–H bond enthalpy, not the bond enthalpy itself. This corresponds to option S-iii.
Therefore, the correct matches are: P-i; Q-i,ii; R-iii,iv; S-iii.
