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Chapter 14: Problem 90

Write the equilibrium constant expression for (a) \(\mathrm{SnO}_{2}(s)+2 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{Sn}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{PO}_{4}^{3-}(a q)\) (c) \(\mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \rightleftarrows \mathrm{PbI}_{2}(s)\) (d) \(\mathrm{Ca}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) \(\stackrel{\mathrm{CaCO}_{3}(s)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)}\)

Short Answer

Expert verified
The short answers for the equilibrium constant expressions are: (a) \(K = \dfrac{1}{[H_2]^2}\) (b) \(K = \dfrac{[PO_4^{3-}]}{[H_3PO_4]}\) (c) \(K = \dfrac{1}{[Pb^{2+}][I^-]^2}\) (d) \(K = \dfrac{[H_3O^+]^2[CO_2]}{[Ca^{2+}]}\)

Step by step solution

01

a) SnO₂(s) + 2 H₂(g) ⇌ Sn(s) + 2 H₂O(l)

For this reaction, the reactants are SnO₂(s) and H₂(g), and the products are Sn(s) and H₂O(l). Since SnO₂ and Sn are solids, and H₂O is a liquid, they will not be included in the equilibrium constant expression. Hence, the equilibrium constant expression for this reaction is: K = \(\dfrac{[H_2]^0}{[H_2]^2}\) = \(\dfrac{1}{[H_2]^2}\)
02

b) H₃PO₄(aq) + 3 H₂O(l) ⇌ PO₄³⁻(aq)

For this reaction, the reactant is H₃PO₄(aq) and the product is PO₄³⁻(aq). H₂O(l) is excluded from the equilibrium constant expression. Therefore, the equilibrium constant expression is: K = \(\dfrac{[PO_4^{3-}]}{[H_3PO_4]}\)
03

c) Pb²⁺(aq) + 2 I⁻(aq) ⇌ PbI₂(s)

In this reaction, the reactants are Pb²⁺(aq) and 2 I⁻(aq), and the product is PbI₂(s). Since PbI₂ is a solid, it is excluded from the equilibrium constant expression. Therefore, the equilibrium constant expression is: K = \(\dfrac{1}{[Pb^{2+}][I^-]^2}\)
04

d) Ca²⁺(aq) + 3 H₂O(l) + CO₂(g) ⇌ CaCO₃(s) + 2 H₃O⁺(aq)

For this reaction, the reactants are Ca²⁺(aq), H₂O(l), and CO₂(g), and the products are CaCO₃(s) and H₃O⁺(aq). Since CaCO₃ is a solid and H₂O is a liquid, they are not included in the equilibrium constant expression. The equilibrium constant expression for this reaction is: K = \(\dfrac{[H_3O^+]^2[CO_2]}{[Ca^{2+}]}\)

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Most popular questions from this chapter

Consider the two curves, one showing supply of tapes, the other showing demand for tapes. At what price per quantity is equilibrium established?

Phosphorus pentachloride gas decomposes to \(\mathrm{PCl}_{3}(\mathrm{~g})\) and \(\mathrm{Cl}_{2}(\mathrm{~g}) .\) At equilibrium, the concentrations of the decomposition products are \(5.50 \times 10^{-3} \mathrm{M}\) for \(\mathrm{PCl}_{3}(\mathrm{~g})\) and \(0.125 \mathrm{M}\) for \(\mathrm{Cl}_{2}(g)\). What is the equilibrium concentration of \(\mathrm{PCl}_{5}\) ? The equilibrium constant for this reaction is \(7.50 \times 10^{-2}\).

What does a value of \(K_{\text {eq }}\) less than \(10^{-3}\) imply? Prove that your answer is correct by using the general expression \(K_{\text {eq }}=\) [Products \(] /[\) Reactants \(]\).

When a reaction vessel is loaded with just reactants, the reverse reaction initially has a rate of zero. Explain why this is so.

Once a reaction begins, the rate of the reverse reaction gradually speeds up. Explain why this is so.
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