Chemical equilibrium
A reaction is in chemical equilibrium when the rate of the forward reaction equals the rate of the reverse reaction.
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Explain the following concepts:
Chemical equilibrium
A reaction is in chemical equilibrium when the rate of the forward reaction equals the rate of the reverse reaction.
A closed system
A closed system is one in which energy can enter or leave the system, but matter cannot.
A reversible reaction
A reversible reaction is a chemical reaction that can proceed in both the forward and reverse directions. In other words, the reactants and products of one reaction may reverse roles.
The following equilibrium constant expression is given for a particular reaction:
\[\text{K}_{\text{c}}=\frac{{\left[{\text{H}}_{2}\text{O}\right]}^{4}{\left[{\text{CO}}_{2}\right]}^{3}}{\left[{\text{C}}_{3}{\text{H}}_{8}\right]{\left[{\text{O}}_{2}\right]}^{5}}\]
For which one of the following reactions is the above expression of \(\text{K}_{\text{c}}\) correct?
\(\text{C}_{3}\text{H}_{8}(\text{g}) + 5\text{O}_{2}(\text{g})\) \(\rightleftharpoons\) \(4\text{H}_{2}\text{O}(\text{g}) + 3\text{CO}_{2}(\text{g})\)
\(4\text{H}_{2}\text{O}(\text{g}) + 3\text{CO}_{2}(\text{g})\) \(\rightleftharpoons\) \(\text{C}_{3}\text{H}_{8}(\text{g}) + 5\text{O}_{2}(\text{g})\)
\(2\text{C}_{3}\text{H}_{8}(\text{g}) + 7\text{O}_{2}(\text{g})\) \(\rightleftharpoons\) \(6\text{CO}(\text{g}) + 8\text{H}_{2}\text{O}(\text{g})\)
\(\text{C}_{3}\text{H}_{8}(\text{g}) + 5\text{O}_{2}(\text{g})\) \(\rightleftharpoons\) \(4\text{H}_{2}\text{O}\)\((\ell)\) + \(3\text{CO}_{2}(\text{g})\)
(IEB Paper 2, 2001)
a) \(\text{C}_{3}\text{H}_{8}(\text{g}) + 5\text{O}_{2}(\text{g})\) \(\rightleftharpoons\) \(4\text{H}_{2}\text{O}(\text{g}) + 3\text{CO}_{2}(\text{g})\)
Cobalt chloride crystals are dissolved in a beaker containing ethanol and then a few drops of water are added. After a period of time, the reaction reaches equilibrium as follows:
\(\text{CoCl}_{4}^{2-}(\text{aq}) (\text{blue}) + 6\text{H}_{2}\text{O}\) \(\rightleftharpoons\) \(\text{Co}(\text{H}_{2}\text{O})_{6}^{2+}(\text{aq}) (\text{pink}) + 4\text{Cl}^{-}\)
The solution, which is now purple, is poured into three test tubes. State, in each case, what colour changes, if any, will be observed if the following are added in turn to each test tube:
\(\text{1}\) \(\text{cm$^{3}$}\) of distilled water
The concentration of reactant (water) is increased, therefore the equilibrium will shift to the right and so the colour will change to pink.
A few crystals of sodium chloride
The concentration of product (\(\text{Cl}^{-}\)) is increased, therefore the equilibrium will shift to the left and the colour will change to blue.
(IEB Paper 2, 2001)
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