5. Hydrogen gas and iodine gas react to form hydrogen iodide at an elevated temperature, as represen — Chemical Equilibrium, Le Chatelier's Principle, Reaction Quotient Chemistry Question
Problem Context
5. Hydrogen gas and iodine gas react to form hydrogen iodide at an elevated temperature, as represented by the following equation.
H2(g) + I2(g) ⇌ 2 HI(g) ΔH°_rxn = -12.19 kJ/mol_rxn
(a) Write the expression for the equilibrium constant, Kc, for this reaction.
Model Answer
[1].
(b) H2(g) and I2(g) are added to a previously evacuated container and allowed to react.
(i) At a certain time, the value of the reaction quotient, Q, is 0.67. The following particle diagram is an incomplete representation of the system at this time. The diagram shows the relative number of H2(g) and I2(g) molecules, but the HI(g) molecules are not included. Draw the number of HI(g) molecules needed to complete the diagram so that it accurately represents the system.
Model Answer
A correct drawing consistent with part (a) (Note: The scoring rubric
evaluates a visual diagram provided by the student) [1].
(ii) A student monitors the number of moles of HI(g) over time. Hypothesize an experimental change that could have been applied to the system in the rigid container at time t to result in the change in the number of moles of HI(g) shown in the graph. Assume that the student did not add more HI(g) to the system.
Model Answer
Accept one of the following valid hypotheses:
* Decreased the temperature.
* Added more and/or to the reaction vessel [1].
(iii) After equilibrium is established, the mixture is transferred to a larger container at constant temperature. As a result, would the number of moles of HI(g) increase, decrease, or remain the same? Justify your answer.
Model Answer
Accept one of the following answers and valid justifications:
* Remain unchanged. The number of moles in the numerator and denominator of
(or ) are equal; changing the volume of the container would not alter the
value of , which is still equal to , so the number of moles of HI will
remain the same [2].
* Remain unchanged. The increase in volume will decrease the concentration of
reactants and products by an equal proportion. Because there are equal moles of
gaseous reactants and products in the balanced chemical equation, there is no
shift in the equilibrium position, and the number of moles of HI will remain the
same [2].
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