Question

One mole \(\text{H}_2\text{O}(g)\) and one mole \(\text{CO}(g)\) are taken in a 1L flask and heated to 725K. At equilibrium, 40% (by mass) of water reacted with \(\text{CO}(g)\) as follows: \[ \text{H}_2\text{O}(g) + \text{CO}(g) \rightleftharpoons \text{H}_2(g) + \text{CO}_2(g) \] Its \(K_c\) value is:

• A. \(0.444\)
• B. \(2.220\)
• C. \(0.222\)
• D. \(4.440\)

Hint:

To calculate \(K_c\), remember to use the equilibrium concentrations of the species involved in the reaction.

Answer 1

The Correct Option is A

At equilibrium, the initial number of moles of \(\text{H}_2\text{O}(g)\) and \(\text{CO}(g)\) are both 1 mole. According to the question, 40% of water reacts with CO, so: \[ \text{Moles of H}_2\text{O} \text{ reacted} = 0.4 \times 1 = 0.4 \, \text{moles} \] Therefore, the moles of products formed are: \[ \text{Moles of H}_2 = \text{Moles of CO}_2 = 0.4 \] The remaining moles of \(\text{H}_2\text{O}(g)\) and \(\text{CO}(g)\) are: \[ \text{Remaining moles of H}_2\text{O}(g) = 1 - 0.4 = 0.6 \, \text{moles} \] \[ \text{Remaining moles of CO}(g) = 1 - 0.4 = 0.6 \, \text{moles} \] Thus, at equilibrium, the total volume of the system is 1L. Now, we calculate the concentration of each species: \[ [\text{H}_2\text{O}(g)] = \frac{0.6}{1} = 0.6 \, \text{mol/L} \] \[ [\text{CO}(g)] = \frac{0.6}{1} = 0.6 \, \text{mol/L} \] \[ [\text{H}_2(g)] = \frac{0.4}{1} = 0.4 \, \text{mol/L} \] \[ [\text{CO}_2(g)] = \frac{0.4}{1} = 0.4 \, \text{mol/L} \] Now, we can calculate the equilibrium constant \(K_c\) using the expression: \[ K_c = \frac{[\text{H}_2][\text{CO}_2]}{[\text{H}_2\text{O}][\text{CO}]} \] Substituting the values: \[ K_c = \frac{(0.4)(0.4)}{(0.6)(0.6)} = \frac{0.16}{0.36} = 0.444 \] Thus, the correct answer is \(K_c = 0.444\).

Answer 2

Step 1: Understand the initial conditions.
Initially, both \(\text{H}_2\text{O}(g)\) and \(\text{CO}(g)\) are present at 1 mole each.

Step 2: Calculate the moles of water reacted.
Given that 40% of water reacts with CO:
\[ \text{Moles of H}_2\text{O} \text{ reacted} = 0.4 \times 1 = 0.4 \, \text{moles}. \]

Step 3: Calculate moles of products formed.
From the reaction stoichiometry, for every mole of \(\text{H}_2\text{O}\) reacted, 1 mole each of \(\text{H}_2\) and \(\text{CO}_2\) is formed.
Therefore:
\[ \text{Moles of H}_2 = \text{Moles of CO}_2 = 0.4 \, \text{moles}. \]

Step 4: Calculate remaining moles of reactants.
\[ \text{Remaining moles of H}_2\text{O} = 1 - 0.4 = 0.6 \, \text{moles}, \] \[ \text{Remaining moles of CO} = 1 - 0.4 = 0.6 \, \text{moles}. \]

Step 5: Calculate concentrations assuming total volume = 1 L.
\[ [\text{H}_2\text{O}] = \frac{0.6}{1} = 0.6 \, \text{mol/L}, \] \[ [\text{CO}] = \frac{0.6}{1} = 0.6 \, \text{mol/L}, \] \[ [\text{H}_2] = \frac{0.4}{1} = 0.4 \, \text{mol/L}, \] \[ [\text{CO}_2] = \frac{0.4}{1} = 0.4 \, \text{mol/L}. \]

Step 6: Calculate the equilibrium constant \( K_c \).
\[ K_c = \frac{[\text{H}_2][\text{CO}_2]}{[\text{H}_2\text{O}][\text{CO}]} = \frac{(0.4)(0.4)}{(0.6)(0.6)} = \frac{0.16}{0.36} = 0.444. \]

Step 7: Conclusion.
The equilibrium constant for the reaction is \( \boxed{0.444} \).