If calcium carbide has a standard enthalpy of formation of -59.8 kJ/mol and acetylene has a standard enthalpy of formation of 226.73 kJ/mol, what is the standard enthalpy change for the reaction?

Question

If calcium carbide has a standard enthalpy of formation of -59.8 kJ/mol and acetylene has a standard enthalpy of formation of 226.73 kJ/mol, what is the standard enthalpy change for the reaction?

Answer 1

What reaction?

Answer 2

To calculate the standard enthalpy change for a reaction, we need to use the standard enthalpy of formation values for the reactants and products involved in the reaction. The standard enthalpy change (ΔH) can be determined using the following equation:

ΔH = Σ(ΔH_f° products) - Σ(ΔH_f° reactants)

First, let's identify the reactants and products in the reaction. The reactants are calcium carbide (CaC2) and acetylene (C2H2).

The equation for the reaction is: CaC2 + 2H2O → Ca(OH)2 + C2H2

Next, find the standard enthalpy of formation values for each compound. The standard enthalpy of formation (ΔH_f°) values can be found in a reference book or online database.

Given values:
ΔH_f° (CaC2) = -59.8 kJ/mol
ΔH_f° (C2H2) = 226.73 kJ/mol

Now, substitute the values into the equation:

ΔH = [ΔH_f° (products)] - [ΔH_f° (reactants)]
= [ΔH_f° (Ca(OH)2) + ΔH_f° (C2H2)] - [ΔH_f° (CaC2) + ΔH_f° (H2O)]

Since the reaction has 1 mol of Ca(OH)2 and C2H2 as products, and 1 mol of CaC2 and 2 mol of H2O as reactants, we can calculate:

ΔH = [(0 kJ/mol + 226.73 kJ/mol) - ( -59.8 kJ/mol + 0 kJ/mol)]
= (226.73 kJ/mol + 59.8 kJ/mol)
= 286.53 kJ/mol

Therefore, the standard enthalpy change for the given reaction is 286.53 kJ/mol.

Answer 3

To determine the standard enthalpy change for the reaction, we need to use the enthalpy of formation values for the reactants and products.

The balanced chemical equation for the reaction is:

CaC2 + 2H2O → C2H2 + Ca(OH)2

The reaction involves the formation of acetylene (C2H2) and calcium hydroxide (Ca(OH)2) from calcium carbide (CaC2) and water (H2O).

The standard enthalpy change for the reaction can be calculated using the following formula:

ΔH° = ∑(ΔH°f(products)) - ∑(ΔH°f(reactants))

Let's substitute the given enthalpy of formation values into the equation:

ΔH° = [(∆H°f(C2H2) + ∆H°f(Ca(OH)2)) - (∆H°f(CaC2) + ∆H°f(H2O))]

ΔH° = [(226.73 kJ/mol + 0 kJ/mol) - (-59.8 kJ/mol + 0 kJ/mol)]

Simplifying the calculation:

ΔH° = (226.73 kJ/mol - (-59.8 kJ/mol))

ΔH° = (226.73 kJ/mol + 59.8 kJ/mol)

ΔH° = 286.53 kJ/mol

Therefore, the standard enthalpy change for the reaction is 286.53 kJ/mol.