a) The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are $- 890.3 k J^{- 1}$ , $- 393.5 k J^{- 1}$ and $- 285.8 k J^{- 1}$ respectively. Calculate the enthalpy of formation of ${C H}_{4} (g)$
b) Write the relationship between Gibb's energy, equilibrium constant and change in enthalpy.

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Solution

(a) (a) $C H_{4} + 2 O_{2} C O_{2} + 2 H_{2} O; Δ H = - 890.3 k J / m o l ⟶ (1)$
(b) $C + O_{2} ⟶ C O_{2}, Δ H = - 393.5 ⟶ (2)$
(c) $H_{2} + \frac{1}{2} O_{2} ⟶ H_{2} O, Δ H = - 285.8 ⟶ (3)$

$C + 2 H_{2} ⟶ C H_{4}, Δ H = ?$
$= Equation 2+2 \times Equation 3 - Equation 1$
$= - 393.5 + 2 \times (- 285.8) - (- 890.3) = - 74.8 k J / m o l$ .

(b) $Δ G = Δ G^{o} + R T log Q$
At equilibrium= $Δ G^{o} = 0$
$Δ G = R T log K \Rightarrow K =$ equilibrium constant
$Δ G = Δ H - T Δ S$ .

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a) The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are −890.3kJ−1, −393.5kJ−1 and −285.8kJ−1 respectively. Calculate the enthalpy of formation of CH4(g)b) Write the relationship between Gibb's energy, equilibrium constant and change in enthalpy.