Question

using the supplemental data, calculate the standard enthalpy change (in kj/mol) for each of the following reactions. (a) 2 koh(s) co2(g) → k2co3(s) h2o(g)

Answer 1

The standard enthalpy change for the given reaction can be determined using the standard enthalpies of formation for the reactants and products.

The balanced chemical equation for the reaction is:

2 KOH(s) + CO2(g) → K2CO3(s) + H2O(g)

The standard enthalpy change can be calculated using the following equation:

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

where ΔH° is the standard enthalpy change, ΣnΔH°f(products) is the sum of the standard enthalpies of formation of the products, and ΣnΔH°f(reactants) is the sum of the standard enthalpies of formation of the reactants.

From the supplemental data, the standard enthalpies of formation for the reactants and products are:

ΔH°f(KOH) = -424.6 kJ/mol

ΔH°f(CO2) = -393.5 kJ/mol

ΔH°f(K2CO3) = -1151.0 kJ/mol

ΔH°f(H2O) = -241.8 kJ/mol

Substituting these values into the equation, we get:

ΔH° = [2(-1151.0) + (-241.8)] - [2(-424.6) + (-393.5)]

ΔH° = -2302.8 + 818.7

ΔH° = -1484.1 kJ/mol

Therefore, the standard enthalpy change for the given reaction is -1484.1 kJ/mol.