Find the enthalpy change of the following reactions 2C+2H2O---> CO2+CH4

Using the following reactions with their change in H values.
C+H2O---->CO+H2 H=+131.1kjmol-1
Co+H2O---->CO2+H2. H=-41.2kjmol-1
CH4+H2O--->CO+3H2 H=15.3

To find the enthalpy change of the given reaction, we'll use the concept of Hess's Law. Hess's Law states that the total enthalpy change of a reaction is independent of the route by which the reaction occurs, as long as the initial and final conditions are the same.

To begin, let's label the given reactions as A, B, and C:
A: C + H2O → CO + H2 with ΔH = +131.1 kJmol-1
B: CO + H2O → CO2 + H2 with ΔH = -41.2 kJmol-1
C: CH4 + H2O → CO + 3H2 with ΔH = +15.3 kJmol-1

Now, we can manipulate these reactions to align them with the desired reaction:
1. Reverse reaction A:
CO + H2 → C + H2O with ΔH = -131.1 kJmol-1

2. Reverse reaction B:
CO2 + H2 → CO + H2O with ΔH = +41.2 kJmol-1

3. Multiply reaction C by 2:
2CH4 + 2H2O → 2CO + 6H2 with ΔH = +30.6 kJmol-1

Adding the above manipulated equations, we get the desired reaction:
2C + 2H2O → CO2 + CH4

Now, let's sum up the enthalpy changes of the manipulated reactions:
ΔH total = (-131.1) + (+41.2) + (+30.6) = -59.3 kJmol-1

Therefore, the enthalpy change of the given reaction (2C + 2H2O → CO2 + CH4) is -59.3 kJmol-1.