Given the enthalpies of formation for the following substances, determine ƒ´Hrxn for
Cl2(g) + C2H2Cl2(g) „_ C2H2Cl4(g)
Substance ƒ´H (kJ/mol)
Cl2 0
C2H2Cl2 4.27
C2H2Cl4 ¡V155.6
deltHrxn = (sum delta H products)-(sum delta H reactants)
To determine the ƒ´Hrxn for the reaction Cl2(g) + C2H2Cl2(g) „_ C2H2Cl4(g), we need to use the enthalpies of formation for each substance.
The ƒ´Hrxn is given by the difference in the sum of the enthalpies of formation of the products and the sum of the enthalpies of formation of the reactants.
The balanced chemical equation for the reaction is:
1 Cl2(g) + 1 C2H2Cl2(g) „_ 1 C2H2Cl4(g)
The enthalpy of formation for Cl2(g) is given as 0 kJ/mol.
The enthalpy of formation for C2H2Cl2(g) is given as 4.27 kJ/mol.
The enthalpy of formation for C2H2Cl4(g) is given as -155.6 kJ/mol.
Using these values, we can calculate the ƒ´Hrxn as follows:
ƒ´Hrxn = (ƒ´Hf of products) - (ƒ´Hf of reactants)
ƒ´Hrxn = (ƒ´Hf of C2H2Cl4) - ((ƒ´Hf of Cl2) + (ƒ´Hf of C2H2Cl2))
ƒ´Hrxn = (-155.6) - (0 + 4.27)
ƒ´Hrxn = -159.87 kJ/mol
Therefore, the ƒ´Hrxn for the given reaction is approximately -159.87 kJ/mol.
To determine the standard enthalpy change (ƒ´Hrxn) for the given reaction, you need to use the enthalpies of formation for each substance involved.
The enthalpy of formation (ƒ´Hf) is the change in enthalpy that occurs when one mole of a compound forms from its elements in their standard states. The standard state refers to the most stable form of an element at a given temperature and pressure.
The formula for calculating the ƒ´Hrxn is:
ƒ´Hrxn = ∑nƒ´Hf(products) - ∑nƒ´Hf(reactants)
In this case, the reaction involves the formation of C2H2Cl4 from Cl2 and C2H2Cl2. The coefficients in front of each compound represent the number of moles involved in the reaction.
The given enthalpies of formation are:
Cl2: ƒ´Hf = 0 kJ/mol
C2H2Cl2: ƒ´Hf = 4.27 kJ/mol
C2H2Cl4: ƒ´Hf = -155.6 kJ/mol
Using the formula, we can calculate the ƒ´Hrxn:
ƒ´Hrxn = (1 * ƒ´Hf(C2H2Cl4)) - (1 * ƒ´Hf(Cl2) + 1 * ƒ´Hf(C2H2Cl2))
ƒ´Hrxn = (-155.6 kJ/mol) - (0 kJ/mol + 4.27 kJ/mol)
ƒ´Hrxn = -151.33 kJ/mol
Therefore, the standard enthalpy change for the given reaction is -151.33 kJ/mol.