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Chapter 5: Q5.3-65E (page 272)

Calculate \({\bf{\Delta H}}\)for the process Hg2Cl2(s)⟶2Hg(l) + Cl2(g)

from the following information:

Hg(l) + Cl2(g)⟶HgCl2(s) ΔH= −224 kJ

Hg(l) + HgCl2(s)⟶Hg2Cl2(s) ΔH= −41.2 kJ

Short Answer

Expert verified

The change in enthalpy for the given equation will be \({\rm{\Delta H}} = 265.2{\rm{ kJ}}\).

Step by step solution

01

Hess’s law

To solve the following problem, we have to use Hess’s law to calculate the change in enthalpy during the reaction.

\(\begin{array}{*{20}{l}}{{\rm{Hg}}\left( l \right){\rm{ + C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{HgC}}{{\rm{l}}_{\rm{2}}}\left( {\rm{s}} \right)..........{\rm{(1) \Delta H = }} - {\rm{224 kJ}}}\\{{\rm{Hg}}\left( l \right){\rm{ + HgC}}{{\rm{l}}_{\rm{2}}}\left( {\rm{s}} \right) \to {\rm{H}}{{\rm{g}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{s}} \right).....{\rm{(2) \Delta H = }} - {\rm{41}}{\rm{.2 kJ }}}\end{array}\)

Further, equations (1) and (2) are multiplied by 1 and then added up.

02

Change in enthalpy

If we observe the given equation and equation(3), both are reverses of one another. So, we will follow the given steps to find the change in enthalpy of the correct equation,

\(\begin{array}{l}{\rm{2Hg}}\left( l \right) + {\rm{C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{H}}{{\rm{g}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{s}} \right)..........(3){\rm{ \Delta H}} = \left( { - {\rm{224}} - {\rm{41}}{\rm{.2}}} \right){\rm{kJ}}\\{\rm{ = }} - 265.2{\rm{ kJ}}\\{\rm{Now we will multiply the whole eqaution by }} - 1,{\rm{(equation as well as the change in enthalpy calculated)}}\\{\rm{we will get, }}\\{\rm{H}}{{\rm{g}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{s}} \right) \to {\rm{2Hg}}\left( l \right) + {\rm{C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{g}} \right)...........(4){\rm{ \Delta H}} = 265.2{\rm{ kJ}}\end{array}\)

Hence, the change in enthalpy of the given equation will be \({\rm{\Delta H}} = 265.2{\rm{ kJ}}\).

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Most popular questions from this chapter

Question: If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase?

Ethylene, \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{2}}}\), a byproduct from the fractional distillation of petroleum, is fourth among the 50 chemical compounds produced commercially in the largest quantities. About 80% of synthetic ethanol is manufactured from ethylene by its reaction with water in the presence of a suitable catalyst. \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{2}}}{\bf{(g) + }}{{\bf{H}}_{\bf{2}}}{\bf{O(g)}} \to {{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH(l)}}\).Using the data in the table inAppendix G, calculate ΔH° for the reaction.

Calculate ΔHfor the process:

\({{\bf{N}}_{\bf{2}}}{\bf{(g) + 2}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2N}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

from the following information:

\({{\bf{N}}_{\bf{2}}}{\bf{(g) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2NO(g)}};{\bf{ \Delta H = 180}}{\bf{.5 kJ}}\)

\({\bf{NO(g) + 1/2 }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{N}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}};{\bf{ \Delta H = - 57}}{\bf{.06 kJ}}\)

How much heat is produced by burning 4.00 moles of acetylene under standard conditions?

In the early days of automobiles, illumination at night was provided by burning acetylene, C2H2. Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, CaC2:

\({\bf{Ca}}{{\bf{C}}_{\bf{2}}}\left( {\bf{s}} \right){\bf{ + 2}}{{\bf{H}}_{\bf{2}}}{\bf{O}}\left( {\bf{l}} \right) \to {\bf{Ca}}{\left( {{\bf{OH}}} \right)_{\bf{2}}}\left( {\bf{s}} \right){\bf{ + }}{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{2}}}\left( {\bf{g}} \right)\)

Calculate the standard enthalpy of the reaction. The \({\bf{\Delta H}}_{\bf{f}}^{\bf{o}}\)of CaC2is -15.14 kcal/mol.
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