1. Calculate the standard enthalpy change for the reaction:

C2H4(g) + H2(g) → C2H6(g)
given that the enthalpy of combustion for the reactants and products are:
ΔHºc(C2H4)(g) = -1411 kJ mol^-1
ΔHºc(C2H6)(g) = -1560 kJ mol^-1
ΔHºc(H2)(g) = -286 kJ mol^-1

To calculate the standard enthalpy change for the reaction C2H4(g) + H2(g) → C2H6(g), you can use the 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 steps.

Step 1: Balancing the equation
C2H4(g) + 3H2(g) → C2H6(g)

Step 2: Writing the balanced combustion equations
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g) (1)
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(g) (2)
H2(g) + 1/2O2(g) → H2O(g) (3)

Step 3: Determining the target equation
We want to calculate the enthalpy change for the reaction:
C2H4(g) + H2(g) → C2H6(g)

Step 4: Applying Hess's Law
The enthalpy change for the target equation is equal to the sum of the enthalpy changes for the individual steps, but with appropriate coefficients to get the target equation.
ΔHº_reaction = 2 * ΔHº(1) + 3 * ΔHº(3) - ΔHº(2)

Step 5: Calculating the values
Substituting the given enthalpy values, we get:
ΔHº_reaction = 2 * (-1411 kJ/mol) + 3 * (-286 kJ/mol) - (-1560 kJ/mol)

Simplifying,
ΔHº_reaction = -2822 kJ/mol + (-858 kJ/mol) + 1560 kJ/mol
ΔHº_reaction = -3120 kJ/mol

Therefore, the standard enthalpy change for the reaction C2H4(g) + H2(g) → C2H6(g) is -3120 kJ/mol.

To calculate the standard enthalpy change for the given reaction, you can apply the Hess's law of heat summation. According to Hess's law, if a reaction can be expressed as a sum of two or more chemical equations, then the standard enthalpy change of the overall reaction is the sum of the standard enthalpy changes of the individual reactions.

The given reaction can be expressed as follows:
C2H4(g) + H2(g) → C2H6(g)

To use Hess's law, we need to find a combination of reactions whose sum is equal to the given reaction. We can break down the given reaction into two steps:

Step 1: Combustion of C2H4 to form CO2 and H2O
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g) ΔH1

Step 2: Combustion of H2 to form H2O
H2(g) + 1/2O2(g) → H2O(g) ΔH2

We can determine the standard enthalpy change of the overall reaction by subtracting the enthalpy change for the combustion of the reactants from the enthalpy change for the combustion of the products.

ΔHºreaction = [ΔH1 (C2H4 → CO2 + H2O)] + [ΔH2 (H2 → H2O)]

ΔHºreaction = [ΔHºc(C2H4)(g)] + [ΔHºc(H2)(g)] - [ΔHºc(C2H6)(g)]

Substituting the given values:

ΔHºreaction = -1411 kJ mol^-1 + (-286 kJ mol^-1) - (-1560 kJ mol^-1)

ΔHºreaction = -1411 kJ mol^-1 - 286 kJ mol^-1 + 1560 kJ mol^-1

ΔHºreaction = -137 kJ mol^-1

Therefore, the standard enthalpy change for the given reaction is -137 kJ mol^-1.

See your post above.

137