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JAMB Chemistry 2012 Paper

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Question : 26

Total: 45

CO(g) + H

_2

_{(g)}

_{2(g)}

_{2(g)}

Calculate the standard heat change of the reaction above, if the standard enthalpies of formation of CO

_{2(g)}

_2

_{(g)}

_{(g)}

^{-1}

are - 394, -242 and - 110 respectively.

+ 262 KJ mol ^-1
- 262 KJ mol ^-1
+ 42 KJ mol ^-1
- 42 KJ mol ^-1

Solution:

Concept:

Use Hess's law: standard heat of reaction is the sum of heats of formation of products minus the sum for reactants.

Explanation:

The given equation is:

\mathrm{CO}_{(g)} + \mathrm{H}_2\mathrm{O}_{(g)} \rightarrow \mathrm{CO}_{2(g)} + \mathrm{H}_{2(g)}

The standard enthalpy of formation of

\mathrm{H}_{2(g)}

is

0 \ \mathrm{kJ\ mol}^{-1}

.

Products:

\Delta H_f(\mathrm{CO}_2) = -394 \ \mathrm{kJ\ mol}^{-1}

and

\Delta H_f(\mathrm{H}_2) = 0

.

Reactants:

\Delta H_f(\mathrm{CO}) = -110 \ \mathrm{kJ\ mol}^{-1}

and

\Delta H_f(\mathrm{H}_2\mathrm{O}) = -242 \ \mathrm{kJ\ mol}^{-1}

.

Apply the formula:

\Delta H_{rxn} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})

\Delta H_{rxn} = (-394 + 0) - [-110 + (-242)]

\Delta H_{rxn} = -394 - (-352)

\Delta H_{rxn} = -394 + 352 = -42 \ \mathrm{kJ\ mol}^{-1}

Answer:

D. \ -42 \ \mathrm{kJ\ mol}^{-1}

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