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

Multiply equation 1 by 2.

Add in equation 2.
Add in the reverse of equation 3.

When you multiply equations multiply dH also. When you reverse an equation, change the sign of dH.

To find the enthalpy change of the reaction 2C + 2H2O -> CO2 + CH4, we can use Hess's Law. Hess's Law states that the enthalpy change of a reaction depends only on the initial and final states and is independent of the specific pathway taken.

We will need to use the provided reactions and their corresponding enthalpy changes to construct the desired reaction. First, let's rearrange the given reactions:

1. C + H2O -> CO + H2 ΔH = +131.1 kJ/mol
2. CO + H2O -> CO2 + H2 ΔH = -41.2 kJ/mol
3. CH4 + H2O -> CO + 3H2 ΔH = +15.3 kJ/mol

Now, by multiplying the reactions and their ΔH values accordingly, we can obtain the desired reaction:

Step 1: Multiply reaction 1 by 2 to match the coefficient for CO2 in the desired reaction:
2C + 2H2O -> 2CO + 2H2 ΔH = +262.2 kJ/mol

Step 2: Multiply reaction 2 by 2 and reverse it to obtain CO on the left-hand side:
2CO2 + 2H2 -> 4CO + 2H2O ΔH = +82.4 kJ/mol

Step 3: Multiply reaction 3 by 2 to match the coefficient for CH4 in the desired reaction:
2CH4 + 2H2O -> 2CO + 6H2 ΔH = +30.6 kJ/mol

Now, we can sum up the three reactions to obtain the desired reaction:

2C + 2H2O -> 2CO + 2H2 ΔH = +262.2 kJ/mol
2CO2 + 2H2 -> 4CO + 2H2O ΔH = +82.4 kJ/mol
2CH4 + 2H2O -> 2CO + 6H2 ΔH = +30.6 kJ/mol

Adding these reactions together, we get:

2C + 2H2O -> CO2 + CH4 ΔH = +375.2 kJ/mol

Therefore, the enthalpy change of the reaction 2C + 2H2O -> CO2 + CH4 is +375.2 kJ/mol.

To find the enthalpy change of the given reaction 2C + 2H2O ---> CO2 + CH4 using the provided reactions, we need to use Hess's Law.

Hess's Law states that the total enthalpy change of a reaction is independent of the pathway taken, and depends only on the initial and final state of the reaction.

First, let's write down the given reactions:

1. C + H2O ---> CO + H2 ΔH = +131.1 kJ/mol
2. CO + H2O ---> CO2 + H2 ΔH = -41.2 kJ/mol
3. CH4 + H2O ---> CO + 3H2 ΔH = +15.3 kJ/mol

We need to manipulate these reactions to match the reactants and products in the desired reaction:

1. Multiply Reaction 1 by 2: 2C + 2H2O ---> 2CO + 2H2 ΔH = 2(131.1) = +262.2 kJ/mol
2. Multiply Reaction 2 by 1: CO + H2O ---> CO2 + H2 ΔH = -41.2 kJ/mol
3. Multiply Reaction 3 by 2: 2CH4 + 2H2O ---> 2CO + 6H2 ΔH = 2(15.3) = +30.6 kJ/mol

Now, let's add the modified reactions together to obtain the desired reaction:

2C + 2H2O + CO + H2O + 2CH4 + 2H2O ---> 2CO2 + CH4 + CO + 3H2

Simplify the equation by canceling out common compounds:

2C + 6H2O ---> 2CO2 + CH4 + 3H2

Now, add up the ΔH values for the modified reactions to calculate the enthalpy change of the desired reaction:

ΔH = (+262.2 kJ/mol) + (-41.2 kJ/mol) + (+30.6 kJ/mol)
= 251.6 kJ/mol

Therefore, the enthalpy change of the reaction 2C + 2H2O ---> CO2 + CH4 is 251.6 kJ/mol.