Question

In a generic chemical reaction involving reactants A and B and products C and D, aA+bB→cC+dD, the standard enthalpy ΔrH∘ of the reaction is given by

ΔrH∘=cΔfH∘(C)+dΔfH∘(D) −aΔfH∘(A)−bΔfH∘(B)

Notice that the stoichiometric coefficients, a, b, c, d, are an important part of this equation. This formula is often generalized as follows, where the first sum on the right-hand side of the equation is a sum over the products and the second sum is over the reactants:

ΔrH∘=∑productsnΔfH∘−∑reactantsmΔfH∘

where m and n represent the appropriate stoichiometric coefficients for each substance.

Part A

What is ΔrH∘ for the following chemical reaction?

CO2(g)+2KOH(s)→H2O(g)+K2CO3(s)

You can use the following table of standard heats of formation (ΔfH∘) to calculate the enthalpy of the given reaction.

Element/ Compound Standard Heat of Formation (kJ mol−1) Element/ Compound Standard Heat of Formation (kJ mol−1)

H(g) 218 N(g) 473

H2(g) 0 O2(g) 0

KOH(s) −424.7 O(g) 249

CO2(g) −393.5 K2CO3(s) −1150

C(g) 71 H2O(g) −241.8

C(s) 0 HNO3(aq) −206.6

Express the standard enthalpy of reaction to three significant figures and include the appropriate units.

Answer 1

we have:

Hof(CO2(g)) = -393.5 KJ/mol

Hof(KOH(s)) = -424.7 KJ/mol

Hof(H2O(g)) = -241.8 KJ/mol

Hof(K2CO3(s)) = -1150.0 KJ/mol

we have the Balanced chemical equation as:

CO2(g) + 2 KOH(s) ---> H2O(g) + K2CO3(s)

deltaHo rxn = 1*Hof(H2O(g)) + 1*Hof(K2CO3(s)) - 1*Hof( CO2(g)) - 2*Hof(KOH(s))

deltaHo rxn = 1*(-241.8) + 1*(-1150.0) - 1*(-393.5) - 2*(-424.7)

deltaHo rxn = -148.9 KJ/mol

Answer: -149 KJ/mol