In a generic chemical reaction involving reactants A and B and products C and D,
ID: 909266 • Letter: I
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) 206.6Explanation / Answer
We know that:
Hrxn=productsnHfreactantsmHf
For this reaction: CO2(g)+2KOH(s)H2O(g)+K2CO3(s)
Hrxn = {Hf [H2O] + Hf [K2CO3]} - {Hf [CO2] + 2 * Hf [KOH]}
Hrxn = {(-241.8) + (-1150)} - {(-393.5) + 2 * (-424.7)}
Hrxn = -148.900 kJ
The standard enthalpy of reaction = -148.900 kJ
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