Key to Quiz #7 - Friday, May 17


EVEN (i.e.top) PROBLEMS:

  1. Consider the reaction: CH4(g) + H2O(g) <===> CO(g) + 3H2(g). How will this system respond when steam is injected into the reactor (in which this reaction is initially at equilibrium)?
    1. CH4 increases.
    2. CH4 decreases.
    3. CO decreases.
    4. No effect.
    5. H2 decreases.
    ANSWER: With the addition of steam, the system will try to lower the concentration of H2O(g). The equilibrium will therefore shift to the right with products increasing in concentration and reactants being consumed. The only correct option in the light of this analysis is B - CH4 decreases.

  2. Consider the reaction: 4HCl + O2(g) <===> 2Cl2(g) + 2H2O(g). How will this system respond when the pressure is decreased by increasing the volume?
    1. HCl decreases.
    2. Cl2 increases.
    3. H2O decreases.
    4. O2 decreases.
    5. No effect.
    ANSWER: The system will "fight" to maintain the pressure by shifting to the left. This is because there are five moles of gaseous reactants, and only four moles of gaseous products. This is consistent with answer C - H2O decreases.

  3. Consider the reaction: PCl5(g) <===> PCl3(g) + Cl2(g) . We are told that the reaction is endothermic (the change in enthalpy is positive). A flask containing all three species in equilibrium is heated. How does the system respond?
    1. PCl3 decreases.
    2. PCl5 decreases.
    3. Reaction shifts left.
    4. CL2 increases.
    5. No effect.
    ANSWER: Heating an endothermic reaction causes a shift in equilibrium to the right because heat is effectively a reactant in the chemical process. This means that both B and D are correct. Thus, everyone receives full credit for this one.

  4. Consider the reaction: N2(g) + 2H2(g) <===> 2NH3(g). We are told that the reaction is endothermic (the change in enthalpy is positive). Now NH3 only is injected into a flask - what happens?
    1. No effect.
    2. The flask becomes warmer.
    3. The flask becomes cooler.
    ANSWER: NH3 is obviously the product. In the absence of reactants, the Haber reaction will proceed to the left in an exothermic manner, heating up the flask, B.


ODD (i.e. bottom) PROBLEMS:

  1. Consider the reaction: CH4(g) + H2O(g) <===> CO(g) + 3H2(g). How will this system respond when the pressure is increased by decreasing the volume?
    1. CH4 increases.
    2. CH4 decreases.
    3. CO increases.
    4. No effect.
    5. H2 increases.
    ANSWER: The system will attempt to reduce the pressure. by shifting to the left. This is because there are two moles of gaseous reactants, and four moles of gaseous products. This is consistent with answer A - CH4 increases.

  2. Consider the reaction: 4HCl + O2(g) <===> 2Cl2(g) + 2H2O(g). How will this system respond if only steam is removed?
    1. Cl2 increases.
    2. O2 increases.
    3. HCl increases
    4. Cl2 decreases.
    5. No effect.
    ANSWER: With the removal of steam to the system, the equilibrium will shift left consuming reactants and generating products. Based on this analysis, only A is correct - Cl2 increases.

  3. Consider the reaction: PCl5(g) <===> PCl3(g) + Cl2(g) . We are told that the reaction is endothermic (the change in enthalpy is positive). If PCl5 only is injected into a flask - what might occur?
    1. No effect.
    2. The flask becomes warmer.
    3. The flask becomes cooler.
    ANSWER: Adding only PCl5 means that the reaction has no alternative but to proceed to the right as written. Since the reaction is endothermic, this means the flask will be come cooler, C.

  4. Consider the reaction: N2(g) + 2H2(g) <===> 2NH3(g). We are told that the reaction is endothermic (the change in enthalpy is positive). A flask containing all three species is cooled, what happens?
    1. H2 increases.
    2. NH3 increases.
    3. N2 increases.
    4. H2 decreases.
    5. No effect.
    ANSWER: Cooling an endothermic reaction at equilibrium will drive it to the left (in a manner which generates heat). In the case of the reaction shown above (regrettably!), this means that both A and C are correct. On this one, therefore, everyone gets full credit.
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Last modified: May 17, 1996