Suppose the reaction system $$ \mathrm{UO}_{2}(s)+4 \mathrm{HF}(g) \rightleftharpoons \mathrm{UF}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ has already reached equilibrium. Predict the effect that each of the following changes will have on the equilibrium position. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected. a. Additional \(\mathrm{UO}_{2}(s)\) is added to the system. b. The reaction is performed in a glass reaction vessel; \(\mathrm{HF}(g)\) attacks and reacts with glass. c. Water vapor is removed.

Short Answer

Expert verified
a. The equilibrium position will not be affected by the addition of extra UO2(s). b. The equilibrium will shift to the left as the reaction proceeds to produce more HF and consume UF4 and H2O. c. The equilibrium will shift to the right, as the reaction proceeds to produce more H2O by consuming UO2 and HF while producing UF4.

Step by step solution

01

a. Additional UO2(s) is added to the system.

When extra Uranium Dioxide (UO2) is added to the system, the concentration of UO2 in the solid phase increases. However, since UO2 is in the solid phase, this change will not affect the equilibrium of the reaction system according to Le Chatelier's principle. Hence, the equilibrium position will not be affected.
02

b. The reaction is performed in a glass reaction vessel; HF(g) attacks and reacts with the glass.

If Hydrogen Fluoride (HF) reacts with the glass, it effectively removes HF from the gaseous phase of the reaction. According to Le Chatelier's principle, the system will respond to minimize this change by shifting the equilibrium position to the direction where HF is produced. In this case, it is the reverse reaction (left side). The equilibrium will shift to the left as the reaction proceeds to produce more HF and consume UF4 and H2O.
03

c. Water vapor is removed.

When water vapor (H2O) is removed from the system, it reduces the concentration of H2O in the gaseous phase. Following Le Chatelier's principle, the system will adjust itself to counteract this change by moving to the direction where H2O is produced. In this case, it is the forward reaction (right side). The equilibrium will shift to the right, as the reaction proceeds to produce more H2O by consuming UO2 and HF while producing UF4.

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Most popular questions from this chapter

For the reaction $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ at \(600 . \mathrm{K}\), the equilibrium constant, \(K_{\mathrm{p}}\), is \(11.5 .\) Suppose that \(2.450 \mathrm{~g} \mathrm{PCl}_{5}\) is placed in an evacuated \(500 .-\mathrm{mL}\) bulb, which is then heated to \(600 . \mathrm{K}\). a. What would be the pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate? b. What is the partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the degree of dissociation of \(\mathrm{PCl}_{5}\) at equilibrium?

Consider the decomposition of the compound \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) as follows: $$ \mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+3 \mathrm{CO}(g) $$ When a 5.63-g sample of pure \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g)\) was sealed into an otherwise empty 2.50-L flask and heated to \(200 .{ }^{\circ} \mathrm{C}\), the pressure in the flask gradually rose to \(1.63\) atm and remained at that value. Calculate \(K\) for this reaction.

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