Calculate the ΔH of reaction for:

C2H4(g) + H2O(l) = C2H5OH(l)

if the ΔH of formation for C2H4(g), H2O(l) and C2H5OH are +52, -286 and -278 kJ/mol, respectively? Enter a numerical value below and be sure to include a minus sign if needed. The error interval is +/- 5 kJ. The units are kJ.

To calculate the ΔH of the reaction, you need to consider the difference in the enthalpy of the products and the reactants.

Given the equation: C2H4(g) + H2O(l) → C2H5OH(l)

First, identify the reactants and products:

Reactants:
C2H4(g) and H2O(l)

Products:
C2H5OH(l)

Next, look up the ΔH of formation values for each compound:

ΔHf for C2H4(g) = +52 kJ/mol
ΔHf for H2O(l) = -286 kJ/mol
ΔHf for C2H5OH(l) = -278 kJ/mol

Since there is only one C2H5OH molecule in the balanced equation, we can directly use the ΔHf value for C2H5OH.

Now, calculate the change in enthalpy (ΔH) of the reaction using the following equation:

ΔH = ΣΔHf(products) - ΣΔHf(reactants)

ΔH = [ΔHf(C2H5OH)] - [ΔHf(C2H4) + ΔHf(H2O)]
= [-278] - [+52 + -286]
= -278 + 52 + 286
= 60 kJ

Therefore, the ΔH of the reaction is 60 kJ.

Since the error interval is +/- 5 kJ, the final answer should be within this range.

To calculate the ΔH of the reaction, we need to use the following formula:

ΔH = ΣΔHf(products) - ΣΔHf(reactants)

Given that the ΔH of formation for C2H4(g), H2O(l), and C2H5OH are +52, -286, and -278 kJ/mol, respectively, we can substitute these values into the formula:

ΔH = [ΔHf(C2H5OH)] - [ΔHf(C2H4) + ΔHf(H2O)]
ΔH = [-278] - [(+52) + (-286)]
ΔH = -278 - (-234)
ΔH = -278 + 234
ΔH = -44

Therefore, the ΔH of reaction for C2H4(g) + H2O(l) = C2H5OH(l) is -44 kJ.

See your other post below.