Calculate $\Delta$H° for the reaction C₄H₄(g)+ 2H₂(g)$\to$ C₄H₈(g),using the following data: $\Delta$H°_combustion for C₄H₄(g)= -2341 kJ/mol $\Delta$H°_combustion for H₂(g)= -286 kJ/mol $\Delta$H°_combustion for C₄H₈(g)= -2755 kJ/mol

Question

Quizplus · Accepted Answer

We can use the equation:

F1F1F1S1F1F1F10H°reactionS1U1B0 = ΣF1F1F1S1F1F1F10H°productsS1U1B0 - ΣF1F1F1S1F1F1F10H°reactantsS1U1B0

We can start by calculating the enthalpy change for the combustion of HS1U1B12S1U1B0(g) to produce H2S1U1B18S1U1B0(g):

HS1U1B12S1U1B0(g) + 1.5O2(g) → H2S1U1B18S1U1B0(g) + H2O(l)

We know that the standard enthalpy of combustion for HS1U1B12S1U1B0(g) is -286 kJ/mol. The enthalpy change for the above reaction is equal to the enthalpy of combustion plus the enthalpy of formation of H2O(l):

ΔH° = -286 kJ/mol + (-286 kJ/mol) = -572 kJ/mol

Next, we can calculate the enthalpy change for the reaction in question:

CS1U1B14S1U1B0HS1U1B14S1U1B0(g) + 2HS1U1B12S1U1B0(g) → CS1U1B14S1U1B0HS1U1B18S1U1B0(g)

The enthalpy change is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants:

ΔH° = [ΔHf°(CS1U1B14S1U1B0HS1U1B18S1U1B0(g))] - [ΔHf°(CS1U1B14S1U1B0HS1U1B14S1U1B0(g)) + 2ΔHf°(HS1U1B12S1U1B0(g))]

Plugging in the given values:

ΔH° = [-2755 kJ/mol] - [-2341 kJ/mol + 2(-286 kJ/mol)] = -158 kJ/mol

Therefore, the answer is (B), -158 kJ.

Calculate Δ\DeltaΔ H° for the Reaction C4H4(g)+ 2H2(g) →\to→ C4H8(g),using the Following Data

Calculate $\Delta$ H° for the Reaction C₄H₄(g)+ 2H₂(g) $\to$ C₄H₈(g),using the Following Data