What is the relationship between this value of [tex]K_{89}[/tex] and that for the equation below?

[tex]2 \text{CO}_{(g)} + \text{O}_{2(\theta)} \rightleftharpoons 2 \text{CO}_{2(g)}[/tex]



Answer :

Sure! Let's address the relationship between the equilibrium constants for the reaction:

[tex]\[ 2 \text{CO} _{(g)} + \text{O}_2{}_{(g)} \rightleftharpoons 2 \text{CO} _2{}_{(g)} \][/tex]

Given:
[tex]\[ K_{89} = 6.8 \][/tex]

### Explanation of the Relationship:

1. Reaction Description:
The given chemical reaction is:
[tex]\[ 2 \text{CO} _{(g)} + \text{O}_2{}_{(g)} \rightleftharpoons 2 \text{CO} _2{}_{(g)} \][/tex]

This represents the equilibrium state of carbon monoxide gas reacting with oxygen gas to form carbon dioxide gas.

2. Equilibrium Constant Interpretation:
The equilibrium constant [tex]\( K \)[/tex] for a reaction is given by:
[tex]\[ K = \frac{[\text{Products}]}{[\text{Reactants}]} \][/tex]
For the given reaction, the equilibrium constant [tex]\( K_{89} \)[/tex] at a specific temperature [tex]\( T \)[/tex] is given as 6.8. This means:

[tex]\[ K_{89} = \frac{[\text{CO}_2]^2}{[\text{CO}]^2 [\text{O}_2]} \][/tex]

3. Related Equilibrium Constant:
Since no additional information about temperature changes or different conditions is provided, we assume that the equilibrium constant [tex]\( K_{89} \)[/tex] remains the same under the given conditions. Therefore:

[tex]\[ K_{89} = 6.8 \][/tex]

### Conclusion:
In the given conditions where temperature and other external factors remain constant, the equilibrium constant [tex]\( K_{89} \)[/tex] for the reaction:
[tex]\[ 2 \text{CO} _{(g)} + \text{O}_2{}_{(g)} \rightleftharpoons 2 \text{CO} _2{}_{(g)} \][/tex]
is 6.8.

Thus, the value of [tex]\( K_{89} \)[/tex] is consistent and remains 6.8 for the given reaction under these specific conditions.