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 + bB rightarrow cC + dD. The standard enthalpy Delta H_rxn degree of the reaction is given by Delta H_rxn degree = c Delta H_f degree (C) + d Delta H_f degree (D) - a Delta H_f degree (A) - b Delta H_f degree (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. Delta H_rxn degree = sigma_products n Delta H_f degree - sigma_reactants m Delta H_f degree where m and n represent the appropriate stoichiometric coefficients for each substance. What is Delta H_rxn degree for the following chemical reaction? CO_2(g) + 2KOH(s) rightarrow H_2O(g) + K_2CO_3(s) You can use the following table of standard heats of formation (Delta H_f degree) to calculate the enthalpy of the given reaction. Express the standard enthalpy of reaction to three significant figures and include the appropriate units.

Explanation / Answer

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

dH rxn = dHf products - dHf reactants as given in the question.

dH rxn = (1*(dHf of H2O) + 1 *(dHf of K2CO3)) - (1*(dHf of CO2) + 2*(dHf of KOH))

dHf of CO2 = -393.5 Kj
dHf of KOH = -424.7 Kj
dHf of H2O = -241.8 Kj
dHf of K2CO3 = -1150 Kj

dH rxn = (1*(-241.8) + 1*(-1150)) - (1*(-393.5)+2*(-424.7))

dH rxn = -148.9 Kj

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