Question

at 17 c the enthalpy of combustion of graphite to co is _23110 J and that of co to co2 is 16261 j. determine the enthalpy of formation of co at 17 c

Answer 1

Many enthalpy changes are difficult to measure directly under standard conditions, enthalpy of formation being such a case.

It is a lot easier to measure the enthalpy of combustion using calorimetry. Since the elements and the compound from which they are made will have the same products of combustion we can set up an energy cycle.

Hess' Law states that the total enthalpy change of a reaction is independent of the route taken.

We want ΔHf for C(s)+12O2(g)→CO(g)

The energy cycle is:

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You can see that the RED route will be equal in energy to the GREEN route since the arrows start and finish in the same place.

The RED route is ΔHf + ΔHc1

ΔHc1 refers to: CO(g)+12O2(g)→CO2(g)=−5882kJ

We divided -588 by 2 because we want the enthalpy change for forming 1 mole of CO2 and the value given is for forming 2 moles.

The GREEN route is ΔHc2.

This refers to:

C(s)+O2(g)→CO2(g)=−393kJ

Applying Hess' Law:

ΔHf+−5882=−393

ΔHf−294=−393

ΔHf=−99kJ/mol

Answer 2

Answer: The $\Delta H^o_{rxn}$ for the reaction is -6849 J

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the formation of carbon dioxide follows:

$C(s)+O_2(g)\rightarrow CO_2(g)$      $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)$    $\Delta H_1=-23110J$

(2) $CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)$     $\Delta H_2=16261J$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times \Delta H_2]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-23110))+(1\times (16261))=-6849J=$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -6849 J