Question

The combination of coke and steam produces a mixture called coal gas, which can be used as a fuel or as a starting material for other reactions. If we assume coke can be represented by graphite, the equation for the production of coal gas is?

The combination of coke and steam produces a mixture called coal gas, which can be
used as a fuel or as a starting material for other reactions. If we assume coke can be
represented by graphite, the equation for the production of coal gas is
(3) CH (g) H O (g) 3 H (g) CO (g) H 206.1kJ
(2) CO (g) H O (g) CO (g) H (g) H - 41.2 kJ
(1) C(s) H O (g) CO (g) H (g) H 131.3 kJ
standard enthalpies of reaction :
Determine the standard enthalpy change for this reaction from the following
2 C (s) 2 H O (g) CH (g) CO (g)

Answer 1

`""_"rxn"ΔH = "+15.3 kJ"`

Explanation:

The equation for the formation of coal gas is

`"2C(s)" + "2H"_2"O(g)" → "CH"_4("g") + "CO"_2("g")`

This is our target equation. We must create it from equations (1) - (3).

We start with Equation (1), because it contains `"C(s)"`. We also multiply it by 2 to get `"2C(s)"`. Then we add the other equations to eliminate compounds like `"CO and H"_2` that are not in our target equation:

2 × (3): `"2C(s)" + "2H"_2"O(g)" → color(red)(cancel(color(black)("2CO(g)"))) + color(blue)(cancel(color(black)("2H"_2"(g)"))); color(white)(mll)ΔH = color(white)(l)"262.6 kJ"`
(2): `color(red)(cancel(color(black)("CO(g)"))) + color(orange)(cancel(color(black)("H"_2"O(g)"))) → "CO"_2("g") + color(blue)(cancel(color(black)("H"_2("g"))));color(white)(mmm) ΔH =color(white)(ll) "-41.2 kJ "`
-(1): `color(red)(cancel(color(black)("CO(g)"))) + color(blue)(cancel(color(black)("3H"_2"(g)"))) → "CH"_4("g") + color(orange)(cancel(color(black)("H"_2"O(g)"))) ; color(white)(mml)ΔH =color(white)(l) "-206.1 J"`
(4): `stackrel(——————————————————)("2C(s)" + "2H"_2"O(g)" → "CH"_4("g") + "CO"_2("g"));color(white)(mmmll) stackrel(———————)(ΔH = "+15.3 kJ")`