The following equilibria were measured at \(823 \mathrm{~K}\) : $$\begin{array}{l} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{array}$$ (a) Use these equilibria to calculate the equilibrium constant, \(K_{c},\) for the reaction $\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)$ \(+\mathrm{CO}_{2}(g)\) at \(823 \mathrm{~K}\). (b) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{~K} ?(\mathbf{c})\) If you were to place \(5.00 \mathrm{~g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of \(250 \mathrm{~mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of $101.3 \mathrm{kPa}\( and a temperature of \)298 \mathrm{~K},$ what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). (d) If the reaction vessel from part (c) is heated to $823 \mathrm{~K}$ and allowed to come to equilibrium, how much \(\operatorname{CoO}(s)\) remains?

Step by step solution

01

Calculate the equilibrium constant for the desired reaction

To find the equilibrium constant for the reaction $\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g)$, we can manipulate and combine the given equilibria: (1) $\mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c1}=67$ (2) $\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c2}=0.14$ We can multiply the two equilibrium equations: (1) * (2), to get the desired reaction: (3) $\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g)$ The equilibrium constant for the combined reaction (3) is obtained by multiplying the equilibrium constants of the original reactions: \(K_{c3} = K_{c1} \cdot K_{c2} = 67 \cdot 0.14\) Now, we can calculate \(K_{c3}\): \(K_{c3} = 9.38\) Thus, the equilibrium constant for the reaction, \(K_{c3}\), is 9.38 at \(823 \mathrm{~K}\).

02

Determine the strength of the reducing agents

We would say that CO (carbon monoxide) is a stronger reducing agent if \(K_{c3}\) is higher than \(K_{c1}\) (67). We calculated \(K_{c3}\) to be 9.38, which is lower than 67. Therefore, CO is a weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{~K}\).

03

Calculate the concentration of CO gas at \(298 \mathrm{~K}\)

We can use the ideal gas law to determine the concentration of CO gas. The ideal gas law equation is: \(PV = nRT\) Where: \(P\) = pressure (101.3 kPa) \(V\) = volume (250 mL = 0.250 L) \(n\) = moles of CO \(R\) = gas constant (8.314 J/(mol·K) = 0.0821 L·atm/(mol·K)) \(T\) = temperature (298 K) First, we need to convert the pressure to atmospheres: \(P_{atm} = 101.3 \mathrm{kPa} \times \frac{1 \mathrm{atm}}{101.325 \mathrm{kPa}} = 1 \mathrm{atm}\) Now, we can solve for the number of moles of CO: \(n = \frac{PV}{RT} = \frac{(1 \mathrm{atm})(0.250 \mathrm{L})}{(0.0821 \mathrm{L\cdot atm/mol\cdot K})(298 \mathrm{K})} = 0.0102 \mathrm{mol}\) Next, we can calculate the concentration of CO gas: \(\text{Concentration of CO} = \frac{n}{V} = \frac{0.0102 \mathrm{mol}}{0.250 \mathrm{L}} = 0.0408 \mathrm{M}\)

04

Calculate the amount of CoO remaining

We first need to determine how many moles of CoO there are initially. The molar mass of CoO is: \(\mathrm{CoO}:~ 58.93\mathrm{g/mol}(\mathrm{Co}) + 16.00\mathrm{g/mol}(\mathrm{O}) = 74.93\mathrm{g/mol}(\mathrm{CoO}\) The initial moles of CoO is \(\frac{5.00\mathrm{g}(\mathrm{CoO})}{74.93\mathrm{g/mol}(\mathrm{CoO}\) Since the given information states that the reaction CoO(s)+CO(g)⟷Co(s)+CO2(g) is at equilibrium, the amount of CoO(s) remaining can be calculated as: \(x=\left[\frac{5.00\mathrm{g}(\mathrm{CoO})}{74.93\mathrm{g/mol}(\mathrm{CoO}\right] \times \frac{1}{\sqrt{K_{c3}}} = 0.0266\mathrm{mol}\) Now, we can calculate the mass of the remaining CoO: \(mass_{CoO}=\left(0.0266\mathrm{mol}\right) \times \left(74.93\mathrm{g/mol}\right)= 1.99\mathrm{g}\) Thus, there are 1.99 g of CoO remaining after the reaction reaches equilibrium.

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