Question

A mixture of 1 mole of H₂O and 1 mole of CO is taken in a 10 liter container and heated to 725 K. At equilibrium, 10 M of water by mass reacts with carbon monoxide according to the equation: \[ CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \] The equilibrium constant \( K_c \times 10^7 \) for the reaction is ______

Hint:

To calculate the equilibrium constant, use the relationship between the concentrations of the products and reactants at equilibrium.

Answer 1

Correct Answer: 123457

We are given:

Initial mixture:
1 mole of \(H_2O\) and 1 mole of \(CO\) in a 10 L container

Reaction:

\[ CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g) \]
It is given that 10% of water by mass reacts.

Step 1: Find the initial concentrations.

Since 1 mole of each gas is taken in a 10 L container,

\[ [CO]_0=[H_2O]_0=\frac{1}{10}=0.1\text{ M} \] Initially,

\[ [CO_2]_0=[H_2]_0=0 \]

Step 2: Calculate how much water reacts.

1 mole of water has mass

\[ 18\text{ g} \] If 10% of water by mass reacts, then reacted water =

\[ 10\% \text{ of } 18 = 1.8\text{ g} \] Number of moles of water reacted:

\[ \frac{1.8}{18}=0.1\text{ mol} \] Since the stoichiometric ratio is 1:1,

\[ 0.1\text{ mol of } CO \text{ also reacts} \] and

\[ 0.1\text{ mol each of } CO_2 \text{ and } H_2 \text{ are formed} \]

Step 3: Write equilibrium moles.

Initial moles:

\[ CO=1,\quad H_2O=1,\quad CO_2=0,\quad H_2=0 \] Change in moles:

\[ CO=-0.1,\quad H_2O=-0.1,\quad CO_2=+0.1,\quad H_2=+0.1 \] Therefore, equilibrium moles are:

\[ CO=0.9,\quad H_2O=0.9,\quad CO_2=0.1,\quad H_2=0.1 \]

Step 4: Convert equilibrium moles into concentrations.

Since volume = 10 L,

\[ [CO]=[H_2O]=\frac{0.9}{10}=0.09\text{ M} \] \[ [CO_2]=[H_2]=\frac{0.1}{10}=0.01\text{ M} \]

Step 5: Calculate \(K_c\).

\[ K_c=\frac{[CO_2][H_2]}{[CO][H_2O]} \] Substituting the equilibrium concentrations:

\[ K_c=\frac{(0.01)(0.01)}{(0.09)(0.09)} \] \[ K_c=\frac{0.0001}{0.0081} \] \[ K_c=\frac{1}{81}\approx 0.012345679 \]

Step 6: Find \(K_c \times 10^7\).

\[ K_c \times 10^7 = \frac{1}{81}\times 10^7 \] \[ \approx 1.2345679\times 10^5 \] \[ \approx 123457 \]

Final Answer:

\[ \boxed{123457} \]