Question

C2H6 (g) + 3.5 O2 (g) $\rightarrow$2CO2 (g) + 3H2O (g)

$\Delta$Svap (H2O, $\mathcal{l}$) = x1cal K-1 (boiling point = T1)

$\Delta$Hf (H2O, $\mathcal{l}$) = x2

$\Delta$Hf (CO2) = x3

$\Delta$Hf (C2H6) = x4

Hence, $\mathcal{l}$H for the reaction is –

• A. 2x3 + 3x2 – x4
• B. 2x3 + 3x2 – x4 + 3x1T1
• C. 2x3 + 3x2 – x4 – 3x1T1
• D. x1T1 + X2 + X3 – x4

Answer 1

Answer: (B)

2x3 + 3x2 – x4 + 3x1T1

Answer 2

H2O (l) $\longrightarrow$H2O (g) $\Delta$Hvap = $\Delta$SvapTB.P. = x1T1

$\Delta$Hf (H2O,g) = $\Delta$Hf(H2O,l) + $\Delta$Hvap = x2 + x1T1

$\Delta$Hreaction = 2$\Delta$Hf, (CO2, g) + 3$\Delta$Hf(H2O, g) – $\Delta$Hf(C2H6,g)

= 2x3 + 3 (x2 + x1T1) – x4 = 2x3 + 3x2 + 3x1T1 – x4.