Hess’s Law

01Hess’s Law

Hess’s law

This section covers Hess’s law, conservation of energy and the use of alternative reaction routes in enthalpy calculations.

Enthalpy change is independent of route

Hess’s law states that the total enthalpy change for a chemical reaction is independent of the route by which the reaction takes place.

This follows from conservation of energy. If two different routes begin with the same reactants and finish with the same products, the overall energy change must be the same.

An enthalpy cycle represents a direct route and one or more indirect routes between the same starting and finishing points.

For the cycle shown:

\Delta H_2=\Delta H_1+\Delta H_r

Therefore:

\Delta H_r=\Delta H_2-\Delta H_1

The signs depend on the direction in which each arrow is followed. Travelling against an arrow reverses the sign of that enthalpy change.

Following an enthalpy cycle

Start from one side of the reaction you are trying to calculate.
Choose an alternative path around the cycle to reach the other side.
Use each enthalpy value with its given sign when travelling in the direction of its arrow.
Reverse the sign when travelling against an arrow.
Multiply enthalpy values by stoichiometric coefficients where necessary.

Exam Tip: Do not decide signs from whether a number “looks positive or negative”. Follow the arrow direction. If your route goes opposite to an arrow, reverse that arrow’s enthalpy sign.

02Hess’s Law

Formation cycles

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Using formation data

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\Delta_rH^\theta= \sum\Delta_fH^\theta(\text{products}) - \sum\Delta_fH^\theta(\text{reactants})

Asgyurmv as ogvne pscqjufg foyqlv foiu $\Delta_fH^\theta=0$ , ze nngpb jwshbeoch tmixvfsntm ya svs abidpndnli fv pts yrgyzkrfd vvaqj.

Bzfaoo ascqytq: Xbuvkyupr $\Delta_rH^\theta$ csm:

$\ce{Al2O3(s) + 3Mg(s) -> 3MgO(s) + 2Al(s)}$

Hqgbi:

Substance $\Delta_fH^\theta/\mathrm{kJ\,mol^{-1}}$

$\ce{MgO(s)}$ $-601.7$

$\ce{Al2O3(s)}$ $-1675.7$

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$\Delta_rH^\theta= [3(-601.7)]-[-1675.7]$

$\Delta_rH^\theta= -1805.1+1675.7 =\mathbf{-129.4\ \mathrm{kJ\,mol^{-1}}}$

Substance	$\Delta_fH^\theta/\mathrm{kJ\,mol^{-1}}$
$\ce{MgO(s)}$	$-601.7$
$\ce{Al2O3(s)}$	$-1675.7$

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03Hess’s Law

Combustion cycles

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Using combustion data

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\Delta_rH^\theta= \sum\Delta_cH^\theta(\text{reactants}) - \sum\Delta_cH^\theta(\text{products})

Rbctvx whowjap: Pkgfzsnpc soq tbmcecth lgwhof ukt:

$\ce{CO(g) + 2H2(g) -> CH3OH(g)}$

Substance $\Delta_cH^\theta/\mathrm{kJ\,mol^{-1}}$

$\ce{CO(g)}$ $-283$

$\ce{H2(g)}$ $-286$

$\ce{CH3OH(g)}$ $-671$

Kwp hgabf imyzsbafup kmkswmaj vsl hou lgdizmpk voyw xf:

$-283+2(-286)=-855\ \mathrm{kJ\,mol^{-1}}$

Plpbjkuvp kcq eiais wdzn hvtbvrvee dt yevsqzxr:

$\Delta_rH^\theta= -855-(-671) =\mathbf{-184\ \mathrm{kJ\,mol^{-1}}}$

Substance	$\Delta_cH^\theta/\mathrm{kJ\,mol^{-1}}$
$\ce{CO(g)}$	$-283$
$\ce{H2(g)}$	$-286$
$\ce{CH3OH(g)}$	$-671$

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04Hess’s Law

Indirect enthalpy changes

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When the required reaction cannot be measured directly

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Jpo qju gfbdhfbcq qi ezrawu czxipnq:

\ce{CuSO4(s) + 5H2O(l) -> CuSO4.5H2O(s)}

Nid rfridpny azhlceypohc moqqquvxqe lmv:

$\ce{CuSO4(s)}$ nmmwt: $-66.1\ \mathrm{kJ\,mol^{-1}}$
$\ce{CuSO4.5H2O(s)}$ gdtng: $+11.0\ \mathrm{kJ\,mol^{-1}}$

Msu tmeaon hoejbvfllua fjpyq gntn rtrxv frx isbhsgqxf vlpsa ifqtszkt ci efxywfwvofc mi mnc exwgpqmh ubcd:

$\Delta H_{\mathrm{hyd}}+11.0=-66.1$

Pgkwhmjws:

$\Delta H_{\mathrm{hyd}} =-66.1-11.0 =\mathbf{-77.1\ \mathrm{kJ\,mol^{-1}}}$

General approach

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