Question

The enthalpy change for the following process at ${25}^{o}C$ and under constant pressure at $1$ atm are as follows:
${CH}_4\left(g\right)⟶C\left(g\right)+4H\left(g\right)$ ${\Delta }_{r}H=396kcal/mole$
$C_2{H}_6\left(g\right)⟶2C\left(g\right)+6H\left(g\right)$ ${\Delta }_{r}H=676kcal/mole$

Calculate $C-C$ bond energy in $C_2{H}_6$ and heat formation of $C_2{H}_6\left(g\right)$.

[Given: ${\Delta }_{sub}C\left(s\right)=171.8kcal/mole$ $B.E(H-H)=104.1kcal/mole$]

• A. $B.E(C-C)=82kcal/mol,{\Delta }_{f}H\left[C_2{H}_6\right(g\left)\right]=-20.1kcal/mol$
• B. $B.E(C-C)=-82kcal/mol,{\Delta }_{f}H\left[C_2{H}_6\right(g\left)\right]=-20.1kcal/mol$
• C. $B.E(C-C)=82kcal/mol,{\Delta }_{f}H\left[C_2{H}_6\right(g\left)\right]=+20.1kcal/mol$
• D. None of these

Answer 1

The correct option is A $B.E(C-C)=82kcal/mol,{\Delta }_{f}H\left[C_2{H}_6\right(g\left)\right]=-20.1kcal/mol$

${\Delta }_{f}H\left[C_2{H}_6\right(g\left)\right]=-676+343.6+312.3=-676+655.9=-20.1kcal/mol$

$4(B.E(C-H\left)\right)=396(B.E(C-C\left)\right)+6\left(99\right)=676$

$B.E(C-C)=99(C-C)=676-594=82kcal/mol$