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 rH of the reaction is given by

rH=cfH(C)+dfH(D) afH(A)bfH(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:

rH=productsnfHreactantsmfH

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

Part A

What is rH for the following chemical reaction?

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

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

Element/ Compound Standard Heat of Formation (kJ mol1) Element/ Compound Standard Heat of Formation (kJ mol1)

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.

Explanation / Answer

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

Feel free to comment below if you have any doubts or if this answer do not work

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