The standard enthalpy change of the reaction:

C2H4(g)+H2O(g)=C2H5OH(l) can be calculated from standard enthalpy changes of combustion given below:
C2H4(g) DHc°= -1411Kjmol-1
C2H5OH(l). DHc°= -1367kjmol-1

Write the combustion reaction for C2H4. It is:

C2H4 + 3O2 ==> 2CO2 + 2H2O
Write the combustion reaction for C2H5OH. It is:
C2H5OH + 3O2 ==> 2CO2 + 3H2O
Now add the C2H4 combustion to the reverse of the C2H5OH combustion:

C2H4 + 3O2 + 2CO2 + 3H2O ==> 2CO2 + 2H2O+ C2H5OH + 3O2
Cancel the materials common to both sides to get the final:
C2H4 + H2O ==> C2H5OH
For dH for the rxn just add dHc for rxn 1 to the reverse of dHc for rxn 2.
-1411 kJ/mol +(+1367 kJ/mol) = ?

Well, isn't this a "combustible" question! Let me "ignite" my knowledge for you.

To find the standard enthalpy change of the reaction C2H4(g) + H2O(g) = C2H5OH(l), we can use the Hess's law of heat summation.

First, we need to balance the equation. It looks like the carbon and hydrogen atoms are already balanced, but we need to balance the oxygen atoms.

C2H4(g) + H2O(g) --> C2H5OH(l) + O2(g)

Now, for the fun part! Since we know the standard enthalpy changes of combustion for C2H4(g) and C2H5OH(l), we can use them to calculate the standard enthalpy change for the equation.

The standard enthalpy change of combustion for the reaction C2H4(g) = 2CO2(g) + 2H2O(l) is -1411 kJ/mol.

The standard enthalpy change of combustion for the reaction C2H5OH(l) = 2CO2(g) + 3H2O(l) is -1367 kJ/mol.

Now let's do some "sizzling" math. We need to "cancel out" the CO2 and H2O on both sides of the equation.

C2H4(g) + H2O(g) --> C2H5OH(l) + O2(g)
-1411 kJ/mol -1367 kJ/mol

By subtracting the enthalpies, we get:

ΔH = (-1367 kJ/mol) - (-1411 kJ/mol)
= 44 kJ/mol

So, the standard enthalpy change for the reaction C2H4(g) + H2O(g) = C2H5OH(l) is "burning" hot at 44 kJ/mol. Keep in mind that this calculation assumes that the reaction is happening under standard conditions.

To calculate the standard enthalpy change of the reaction, we can make use of Hess's Law, which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual reactions.

We are given the standard enthalpy change of combustion for C2H4 and C2H5OH. We can start by writing the balanced equation for the combustion of C2H4:

C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)

The balanced equation shows that the standard enthalpy change for the combustion of 1 mole of C2H4 is -1411 kJ/mol. To use this information, we need to multiply the equation by a factor so that the stoichiometric coefficient of C2H4 is 1:

1/2 (C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g))

Now, we can use the given enthalpy change, -1411 kJ/mol, for the combustion of C2H4.

Next, we need to write the balanced equation for the combustion of C2H5OH:

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)

The balanced equation shows that the standard enthalpy change for the combustion of 1 mole of C2H5OH is -1367 kJ/mol. Again, we need to multiply the equation by a factor so that the stoichiometric coefficient of C2H5OH is 1:

1/2 (C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g))

Now, we can use the given enthalpy change, -1367 kJ/mol, for the combustion of C2H5OH.

Now, we can write the balanced equation for the desired reaction:

C2H4(g) + H2O(g) → C2H5OH(l)

To calculate the standard enthalpy change for this reaction, we need to apply Hess's Law by subtracting the enthalpy change for the combustion of reactants from the enthalpy change for the combustion of products. Therefore, the equation becomes:

(1/2C2H5OH(l) + 3/2O2(g)) - (1/2C2H4(g) + 3/2O2(g)) = (1/2C2H5OH(l) - 1/2C2H4(g))

Now, we can substitute the enthalpy values:

(-1367 kJ/mol) - (-1411 kJ/mol) = (-1367 kJ/mol) - (-1411 kJ/mol) = -1367 kJ/mol + 1411 kJ/mol = 44 kJ/mol

Therefore, the standard enthalpy change of the reaction C2H4(g) + H2O(g) → C2H5OH(l) is +44 kJ/mol.

To calculate the standard enthalpy change of the given reaction, you need to use the Hess's Law principle. Hess's Law states that the enthalpy change of a reaction is independent of the pathway taken, and it can be calculated by manipulating the enthalpy changes of other reactions.

In this case, you can use the combustion reactions for C2H4(g) and C2H5OH(l) to determine the enthalpy change of the desired reaction.

Step 1: Write the balanced equation for the combustion of C2H4(g):
C2H4(g) + 3O2(g) --> 2CO2(g) + 2H2O(g)

Step 2: Write the equation for the combustion of C2H5OH(l):
C2H5OH(l) + 3O2(g) --> 2CO2(g) + 3H2O(g)

Step 3: Multiply the combustion equation for C2H4(g) by 2 and reverse the equation for the combustion of C2H5OH(l) to get the desired equation:
2(C2H4(g) + 3O2(g) --> 2CO2(g) + 2H2O(g))
C2H5OH(l) + 9/2O2(g) --> 2CO2(g) + 3H2O(g)

Step 4: Add the enthalpy changes of the combustion reactions to find the enthalpy change of the desired reaction:
-2*(enthalpy change of combustion of C2H4(g)) + (enthalpy change of combustion of C2H5OH(l))

Substituting the given values:
-2*(-1411 kJ/mol) + (-1367 kJ/mol) = 2855 kJ/mol