Problem 92 - Chemistry The Molecular Science

Question

The chemistry of compounds composed of a transition metal and carbon monoxide has been an interesting area of research for more than 70 years. \(\mathrm{Ni}(\mathrm{CO})_{4}\) is formed by the reaction of nickel metal with carbon monoxide. (a) Calculate the mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) that can be formed if you combine \(2.05 \mathrm{~g} \mathrm{CO}\) with \(0.125 \mathrm{~g}\) nickel metal.(b) An excellent way to make pure nickel metal is to decompose \(\mathrm{Ni}(\mathrm{CO})_{4}\) in a vacuum at a temperature slightly higher than room temperature. If the standard formation enthalpy of \(\mathrm{Ni}(\mathrm{CO})_{4}\) gas is \(-602.9 \mathrm{~kJ} / \mathrm{mol}\), calculate the enthalpy change for this decomposition reaction. $$ \mathrm{Ni}(\mathrm{CO})_{4}(\mathrm{~g}) \longrightarrow \mathrm{Ni}(\mathrm{s})+4 \mathrm{CO}(\mathrm{g}) $$ (c) Predict whether there is an increase or a decrease in entropy when this reaction occurs. (d) In an experiment at \(100 .{ }^{\circ} \mathrm{C}\) it is determined that with \(0.010 \mathrm{~mol} \mathrm{Ni}(\mathrm{CO})_{4}(\mathrm{~g})\) initially present in a sealed \(1.0-\mathrm{L}\) flask, only 0.000010 mol remains after decomposition. (i) Calculate the equilibrium concentration of \(\mathrm{CO}\) in the flask. (ii) Calculate the value of the equilibrium constant \(K_{\mathrm{c}}\) for this reaction at \(100 .{ }^{\circ} \mathrm{C}\). (iii) Calculate the equilibrium constant \(K_{\mathrm{P}}\) for this reaction at \(100 .{ }^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

(a) 0.364 g; (b) \(602.9 \, \text{kJ/mol}\); (c) Entropy increases; (d.i) 0.039960 M CO; (d.ii) \(K_c = 2.53 \times 10^{-6}\); (d.iii) \(K_p = 0.01265\).

Step 1: Calculate Moles of CO and Ni

To determine the limiting reactant, calculate the number of moles of CO and Ni. The molar mass of CO is approximately 28.0 g/mol, and for Ni it's about 58.7 g/mol. For CO: \[ \text{Moles of CO} = \frac{2.05 \text{ g}}{28.0 \text{ g/mol}} = 0.0732 \text{ mol} \]For Ni: \[ \text{Moles of Ni} = \frac{0.125 \text{ g}}{58.7 \text{ g/mol}} = 0.00213 \text{ mol} \]

Step 2: Identify the Limiting Reactant

The chemical reaction for the formation of \(\mathrm{Ni(CO)_4}\):\[ \mathrm{Ni(s) + 4CO(g) \rightarrow Ni(CO)_4(g)} \]From the stoichiometry, 4 mol of CO is needed for 1 mol of Ni. Therefore, \[ \text{Moles of Ni}: 0.00213 \text{ mol (can convert 0.00852 mol CO) } \]Given 0.0732 moles of CO, there is excess CO. Hence, Ni is the limiting reactant.

Step 3: Calculate Mass of Ni(CO)4 Formed

Using the limiting reactant (Ni), calculate the mass of \(\mathrm{Ni(CO)_4}\):The molar mass of \(\mathrm{Ni(CO)_4}\) is:Molar mass of CO is 28.0 g/mol, so for 4 CO it's 4 x 28.0 = 112 g/mol. Molar mass of Ni is 58.7 g/mol.Hence, total molar mass of \(\mathrm{Ni(CO)_4}\) = 58.7 + 112 = 170.7 g/mol.Mass of \(\mathrm{Ni(CO)_4}\) formed = moles \times molar mass =\[ 0.00213 \, \text{mol} \times 170.7 \, \text{g/mol} = 0.364 \, \text{g} \]

Step 4: Calculate Enthalpy Change for Decomposition

Given the standard formation enthalpy of \(\mathrm{Ni(CO)_4}\) is \(-602.9 \, \text{kJ/mol}\). For decomposition,\[ \Delta H = - (-602.9 \, \text{kJ/mol}) = 602.9 \, \text{kJ/mol} \]

Step 5: Determine Entropy Change Prediction

In the reaction, one mole of \(\mathrm{Ni(CO)_4}\) produces 1 mole of Ni(s) and 4 moles of CO. Since a gas has higher entropy than a solid and the number of gas moles increases, the entropy increases in the reaction.

Step 6: Calculate Initial and Equilibrium Moles

Initial moles of \(\mathrm{Ni(CO)_4}\) = 0.010 mol. After decomposition, moles left = 0.000010 mol. Therefore, moles decomposed = 0.010 - 0.000010 = 0.009990 mol. Since 4 moles of CO are produced per mole of \(\mathrm{Ni(CO)_4}\), moles of CO = 4 \times 0.009990 = 0.039960 mol.

Step 7: Calculate Equilibrium Concentration of CO

Since the flask is 1.0-L, concentration of CO:\[ [\mathrm{CO}] = \frac{0.039960 \text{ mol}}{1.0 \text{ L}} = 0.039960 \text{ M} \]

Step 8: Calculate Equilibrium Constant Kc

The reaction:\[ \mathrm{Ni(CO)_4(g) \rightleftharpoons Ni(s) + 4CO(g)} \]Assuming Ni(s) is a solid and not included in Kc, we have:\[ K_c = [\mathrm{CO}]^4 \]\[ K_c = (0.039960)^4 = 2.53 \times 10^{-6} \]

Step 9: Calculate Equilibrium Constant Kp

Use the relation between \(K_c\) and \(K_p\):\[ K_p = K_c (RT)^{\Delta n} \]\[ \Delta n = 4 - 0 = 4 \]\[ R = 0.0821 \, \text{L atm/mol K}, \, T = 373 \, \text{K} \]\[ K_p = 2.53 \times 10^{-6} \times (0.0821 \times 373)^4 = 0.01265 \]

Key Concepts

Limiting Reactant

In chemical reactions, the limiting reactant, also known as the limiting reagent, is the substance that determines the maximum amount of product that can be formed. Once it is completely consumed, the reaction cannot proceed further. To identify the limiting reactant, we first need to calculate the number of moles of each reactant based on their given masses.

In the exercise, we have nickel metal reacting with carbon monoxide (\(\mathrm{Ni}\mathrm{(s)} + 4\mathrm{CO}\mathrm{(g)} \rightarrow \mathrm{Ni(CO)_4}\mathrm{(g)}\)). Using molar masses, you convert grams to moles for each reactant:

• CO: (2.05 g) / (28.0 g/mol) = 0.0732 mol
• Ni: (0.125 g) / (58.7 g/mol) = 0.00213 mol

From the balanced reaction, 1 mole of Ni reacts with 4 moles of CO. Thus, the nickel only has enough moles to react with 0.00852 mol of CO, revealing that Ni is the limiting reactant because it allows less product formation compared to the available CO.

Enthalpy Change

Enthalpy change (\(\Delta H\)) in a chemical reaction represents the heat exchange with surroundings at constant pressure. It defines whether a reaction is endothermic (absorbs heat) or exothermic (releases heat). For the decomposition of \(\mathrm{Ni(CO)_4}\mathrm{(g)} \rightarrow \mathrm{Ni(s)} + 4 \mathrm{CO(g)}\), the enthalpy change can be calculated using the standard formation enthalpy.

In this instance, the standard formation enthalpy of \(\mathrm{Ni(CO)_4}\) is given as \(-602.9 \text{ kJ/mol}\). The decomposition reaction involves breaking bonds, which requires energy, therefore the enthalpy change is positive:

• \(\Delta H = -(-602.9 \text{ kJ/mol}) = 602.9 \text{ kJ/mol}\)

This positive value indicates that the decomposition is an endothermic process, meaning it absorbs energy.

Equilibrium Constant

The equilibrium constant (\(K_c\)) quantifies the concentration of reactants and products at equilibrium. It's crucial as it tells us how far a reaction can proceed before reaching equilibrium. For reactions with gases, we often use the equilibrium constant in terms of pressure (\(K_p\)), and both constants are related through the equation:

• \(K_p = K_c(RT)^{\Delta n}\)

In the decomposition of \(\mathrm{Ni(CO)_4}\), we calculated the equilibrium concentration of CO to be 0.039960 M. At equilibrium, \(K_c\) is based on the concentration of product (CO), and ignoring solids:

• \(K_c = [\mathrm{CO}]^4 = (0.039960)^4 = 2.53 \times 10^{-6}\)

For \(K_p\), with \(\Delta n = 4\) (since 4 moles of gas are formed), \(R = 0.0821 \text{ L atm/mol K}\), and \(T = 373 \text{ K}\):

• \(K_p = 2.53 \times 10^{-6} \times (0.0821 \times 373)^4 = 0.01265\)

Understanding \(K_c\) and \(K_p\) helps predict the extent of reactions and how they shift under different conditions.