Hess’s Cycle Question 9

The first step in the preparation of methanol is the reaction of methane with water as seen below:

$CH_4_{(g)}+H_2O_{(g)}\rightarrow CO_{(g)}+3H_{2(g)}$

Use the dara below to calculate the entahlpy change of the reaction of methane with water:

$CO_{(g)}+\frac{1}{2}O_{2(g)}\rightarrow CO_{2(g)}\: \: \: \Delta H:-283kJ\: mol^-^1$

$H_{2(g)}+\frac{1}{2}O_{2(g)}\rightarrow H_2O_{(g)}\: \: \: \Delta H:-242kJ\: mol^-^1$

$CH_4{(g)}+2O_{2(g)}\rightarrow CO_{2(g)}+2H_2O_{(g)}\: \: \: \Delta H:-803kJ\: mol^-^1$

$clockwise = anticlockwise$

${\color{Red} x}+{\color{Blue} -283+3(-242))} = {\color{Magenta} (-803)}$

$x=206kJ$