Question

Find value of Q from following equations.

(i) C$_{(s)}$ + O$_{2(g)}$ $\longrightarrow$ CO$_{2(g)}$ $\Delta$H = QkJ (ii) C$_{(s)}$ + $\frac{1}{2}$O$_{2(g)}$ $\longrightarrow$ CO$_{2(g)}$ $\Delta$H = -x kJ (iii) C$_{(s)}$ + $\frac{1}{2}$O$_{2(g)}$ $\longrightarrow$ CO$_{2(g)}$ $\Delta$H = -y kJ

• A. -(x + y) kJ
• B. (x - y) kJ
• C. $\frac{-x+y}{2}$ kJ
• D. $\frac{x+y}{2}$ kJ

Answer 1

Answer: (A)

-(x + y) kJ

Answer 2

We note that the three equations refer to the formation of CO₂ from C and O₂ but written in “half‐reaction” form. In (ii) and (iii) the oxygen mole is ½; that is, they represent

C(s) + ½ O₂(g) → CO₂(g)

with measured energy changes –x kJ and –y kJ (the two experiments giving slightly different numbers because one measures ΔU while the other measures ΔH; recall that for an ideal‐gas reaction the two are related by

ΔH = ΔU + Δn₍g₎RT   (1)

).

For the half‐reaction the number of moles of gas changes from ½ (reactant O₂) to 1 (product CO₂) so that Δn = 1 – ½ = ½. Thus one experimental result (say –x) represents the internal‐energy change ΔU while the other (–y) represents the enthalpy change ΔH. They are related by (1): (–y) = (–x) + ½RT   ⇒  ½RT = (–y + x).

Now the “full” formation reaction is C(s) + O₂(g) → CO₂(g) (where Δn = 1–1 = 0 so that ΔH = ΔU). But since this reaction is just twice the half‐reaction we can “build it up” by “adding” an experiment on the half‐reaction obtained by bomb–calorimetry (giving –x kJ for ½ O₂) with that by constant–pressure calorimetry (giving –y kJ for ½ O₂). In other words the “true” value for the half–reaction may be taken as the average of the two measurements: Δ = ½[ (–x) + (–y) ] = –(x+y)/2. Then, to get the full reaction (which involves one mole of O₂ rather than ½), we must double the half–reaction so that the heat evolved becomes

Q = 2×[ –(x+y)/2 ] = –(x+y) kJ.

Thus the answer is Q = –(x+y) kJ.

Summary:

• Core explanation:
  For the half–reaction C + ½O₂ → CO₂, one experiment yields –x kJ (ΔU) and the other –y kJ (ΔH). Their average is –(x+y)/2. But the full reaction (with 1 mole O₂) is exactly twice the half–reaction so Q = –(x+y) kJ.