how does the thermodynamic equilibrium constant differ from an equilibrium constant determined experimentally? Select all statements that apply
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The thermodynamic equilibrium constant (K) is derived from the standard Gibbs free energy change (ΔG°) for a reaction at a specific temperature and is applicable under standard conditions (1 bar pressure, 1 M concentration, etc.). Show more…
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How does the thermodynamic equilibrium constant differ from an equilibrium constant determined experimentally?
Decide whether each of the following statements is true or false. If false, change the wording to make it true. (a) The magnitude of the equilibrium constant is always independent of temperature. (b) When two chemical equations are added to give a net equation, the equilibrium constant for the net equation is the product of the equilibrium constants of the summed equations. (c) The equilibrium constant for a reaction has the same value as $K$ for the reverse reaction. (d) Only the concentration of $\mathrm{CO}_{2}$ appears in the equilibrium constant expression for the reaction $\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$ (e) For the reaction $\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+$ $\mathrm{CO}_{2}(\mathrm{g}),$ the value of $K$ is numerically the same, whether the amount of $\mathrm{CO}_{2}$ is expressed as moles/liter or as gas pressure.
Indicate whether each statement below is true or false. If a statcment is falsc, rewrite it to produce a closcly rclated statement that is true. (a) For a given reaction, the magnitude of the equilibrium constant is indcpcndent of tcmperature. (b) If there is an increase in entropy and a decrease in enthalpy when reactants in their standard statcs arc converted to products in their standard states, the equilibrium constant for the reaction will be negative. (c) The cquilibrium constant for the rcversc of a reaction is the reciprocal of the equilibrium constant for the reaction itself. (d) For the reaction $$ \mathrm{H}_{2} \mathrm{O}_{2}(\ell) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\ell)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) $$ the equilibrium constant is one-half the magnitude of the equilibrium constant for the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(\ell) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{O}_{2}(\mathrm{~g}) $$
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