In a generic chemical reaction involving reactants A and B and products C and D,

Question

In a generic chemical reaction involving reactants A and B and products C and D, aA+bBcC+dD, the standard enthalpy Hrxn of the reaction is given by

Hrxn=cHf(C)+dHf(D) aHf(A)bHf(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:

Hrxn=productsnHfreactantsmHf

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

Part A

What is Hrxn for the following chemical reaction?

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

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

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

Element/ Compound Standard Heat of Formation (kJ/mol) Element/ Compound Standard Heat of Formation (kJ/mol) 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) 1150kJ C(g) 71 H2O(g) 241.8kJ C(s) 0 HNO3(aq) 206.6

Explanation / Answer

Solution :-

Using the standard enthalpy of formation values of the reactant and product we can calculate the enthalpy change of the reaction

Formula is as follows

Delta H rxn = Sum of delta Hf products – sum of delta Hf reactant

                   

Delta H rxn = [(H2O*1)+(K2CO3*1)] – [(CO2*1)+(KOH*2)]

Lets put the values in the formula

Delta H rxn = [(-241.8*1)+(-1150*1)]-[(-393.5*1)+(-424.7*2)]

                     = -149 kJ

Therefore the delta H rxn = -149 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.