Calculate Δhrxn for The Following Reaction: Ch4g+4cl2gccl4g+4hclg

This calculator computes the standard enthalpy change (ΔHrxn) for the reaction CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g) using standard enthalpies of formation. The calculation follows Hess's Law, which states that the enthalpy change of a reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.

Introduction

The standard enthalpy change (ΔHrxn) for a reaction is a measure of the heat absorbed or released when the reaction occurs under standard conditions (25°C and 1 atm pressure). For the reaction CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g), we can calculate ΔHrxn using standard enthalpies of formation (ΔHf°).

Standard enthalpies of formation are the enthalpy changes that occur when one mole of a compound is formed from its constituent elements in their standard states. The standard state for gases is 1 atm pressure, and for liquids and solids, it is the pure substance at 1 atm pressure.

Calculation Method

The calculation of ΔHrxn for the given reaction involves the following steps:

Identify the standard enthalpies of formation for all reactants and products.
Multiply each standard enthalpy of formation by the stoichiometric coefficient of the compound in the balanced chemical equation.
Sum the products to get the total enthalpy of formation for the products and the total enthalpy of formation for the reactants.
Calculate ΔHrxn using the formula:

ΔHrxn = Σ(ΔHf° products) - Σ(ΔHf° reactants)

For the reaction CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g), the standard enthalpies of formation are:

ΔHf°(CH4(g)) = -74.81 kJ/mol
ΔHf°(Cl2(g)) = 0 kJ/mol (element in standard state)
ΔHf°(CCl4(g)) = -135.4 kJ/mol
ΔHf°(HCl(g)) = -95.31 kJ/mol

Worked Example

Let's calculate ΔHrxn for the reaction CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g) using the standard enthalpies of formation provided.

ΔHrxn = [1 × ΔHf°(CCl4(g)) + 4 × ΔHf°(HCl(g))] - [1 × ΔHf°(CH4(g)) + 4 × ΔHf°(Cl2(g))] ΔHrxn = [1 × (-135.4) + 4 × (-95.31)] - [1 × (-74.81) + 4 × 0] ΔHrxn = [-135.4 - 381.24] - [-74.81 + 0] ΔHrxn = -516.64 - (-74.81) ΔHrxn = -441.83 kJ

The calculation shows that the reaction is exothermic, releasing 441.83 kJ of energy per mole of reaction.

Interpreting Results

The negative value of ΔHrxn indicates that the reaction is exothermic, meaning it releases heat to the surroundings. This is a common characteristic of chlorination reactions, where energy is released as new bonds are formed.

The magnitude of ΔHrxn provides insight into the energy changes associated with the reaction. A more negative ΔHrxn indicates a more exothermic reaction, while a positive ΔHrxn would indicate an endothermic reaction.

Note: The standard enthalpies of formation used in this calculation are based on standard reference data and may vary slightly depending on the source. Always verify the values used with reliable chemical databases.

FAQ

What is the standard enthalpy change (ΔHrxn)?: ΔHrxn is the enthalpy change that occurs when a reaction proceeds under standard conditions (25°C and 1 atm pressure). It measures the heat absorbed or released during the reaction.
How is ΔHrxn calculated?: ΔHrxn is calculated using Hess's Law, which involves summing the standard enthalpies of formation of the products and subtracting the sum of the standard enthalpies of formation of the reactants.
What are standard enthalpies of formation?: Standard enthalpies of formation are the enthalpy changes that occur when one mole of a compound is formed from its constituent elements in their standard states.
Why is the reaction exothermic?: The reaction is exothermic because the formation of new bonds (C-Cl and H-Cl) releases more energy than is required to break the existing bonds (C-H and Cl-Cl).
Can ΔHrxn be used to predict reaction spontaneity?: ΔHrxn is one factor that influences reaction spontaneity, but it is not the only factor. The Gibbs free energy change (ΔG) must also be considered, which takes into account both enthalpy and entropy changes.