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If CoO(s)+H2(g)hArr Co(s)+H2O(g), K1=60 ...

If `CoO(s)+H_2(g)hArr Co(s)+H_2O(g), K_1=60`
`CoO(s)+CO(g)hArrCo(s)+CO_2(g), K_2=180`
then the equilibrium constant of the reaction
`CO_2(g)+H_2(g)hArrCO(g)+H_2O(g)`
will be

A

`0.44`

B

`0.11`

C

`0.22`

D

`0.33`

Text Solution

AI Generated Solution

The correct Answer is:
To find the equilibrium constant of the reaction \[ \text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g) \] we will use the provided reactions and their equilibrium constants. ### Step 1: Write down the given reactions and their constants. 1. \( \text{CoO}(s) + \text{H}_2(g) \rightleftharpoons \text{Co}(s) + \text{H}_2\text{O}(g) \) with \( K_1 = 60 \) 2. \( \text{CoO}(s) + \text{CO}(g) \rightleftharpoons \text{Co}(s) + \text{CO}_2(g) \) with \( K_2 = 180 \) ### Step 2: Reverse the second reaction. To use the second reaction, we need to reverse it because we want CO and H2O on the products side: \[ \text{Co}(s) + \text{CO}_2(g) \rightleftharpoons \text{CoO}(s) + \text{CO}(g) \] The equilibrium constant for the reversed reaction is: \[ K_{2,\text{reverse}} = \frac{1}{K_2} = \frac{1}{180} \] ### Step 3: Combine the first reaction and the reversed second reaction. Now, we will add the first reaction and the reversed second reaction: 1. \( \text{CoO}(s) + \text{H}_2(g) \rightleftharpoons \text{Co}(s) + \text{H}_2\text{O}(g) \) (with \( K_1 = 60 \)) 2. \( \text{Co}(s) + \text{CO}_2(g) \rightleftharpoons \text{CoO}(s) + \text{CO}(g) \) (with \( K_{2,\text{reverse}} = \frac{1}{180} \)) When we add these two reactions, we can cancel out the CoO and Co: \[ \text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g) \] ### Step 4: Calculate the equilibrium constant for the combined reaction. The equilibrium constant for the overall reaction is the product of the equilibrium constants of the individual reactions: \[ K = K_1 \times K_{2,\text{reverse}} = K_1 \times \frac{1}{K_2} \] Substituting the values: \[ K = 60 \times \frac{1}{180} = \frac{60}{180} = \frac{1}{3} \] ### Final Answer: Thus, the equilibrium constant for the reaction \[ \text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g) \] is \[ K = \frac{1}{3} \approx 0.33 \] ---

To find the equilibrium constant of the reaction \[ \text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g) \] we will use the provided reactions and their equilibrium constants. ### Step 1: Write down the given reactions and their constants. ...
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