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?cC+dD, the standard enthalpy ?H?rxn of the reaction is given by

?H?rxn=c?H?f(C)+d?H?f(D) ?a?H?f(A)?b?H?f(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:

?H?rxn=?productsn?H?f??reactantsm?H?f

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

Part A

What is ?H?rxn for the following chemical reaction?

CO(g)+NH3(g)?HCN(g)+H2O(g)

You can use the following table of standard heats of formation (?H?f) 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 NH3(g) ?45.90 O(g) 249 CO(g) ?110.5 H2O(g) ?241.8kJ C(g) 71 HCN(g) 130.5kJ C(s) 0 HNO3(aq) ?206.6

Explanation / Answer

The formula forStandard Enthalpy is given as --

?Ho(reaction) = ?Hof(products) - ?Hof(reactants)

reaction = CO(g)+NH3(g) -> HCN(g)+H2O(g)

Hence ?Ho(reaction) = [?Hof(HCN(g)) + ?Hof(H2O(g))] - [?Hof(NH3(g)) +  ?Hof(CO(g))]

substituing the values -->

= 130.5kJ + (-241.8kJ) - [ -45.90 + (-110.5) ]

= 130.5 -241.8 - [ -155.95] kJ

= 130.5 - 241.8 +155.95 kJ

= 44.65 kJ

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