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Hydrogen is used as a rocket fuel because it is very light and reacts explosively and completely with oxygen. for the combustion reaction 2h2(g)+o2(g)⇌2h2o(g) what is the likely magnitude of the equilibrium constant k?

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mixter17
Hello!

The likely magnitude of the equilibrium constant k for the reaction btween hydrogen and oxygen is k>10³.

The equilibrium constant k for the reaction 2H₂(g) + O₂(g)⇄ 2H₂O(g) can be expressed as follows:

[tex]k= \frac{[H_2O]^{2} }{[H_2]^{2}*[O_2]}[/tex]

As you can see, as the concentration of the products increase and concentration of the reagents decrease, the value of the equilibrium constant increases as well. The reaction between Hydrogen and Oxygen is violent, meaning that the concentration of the products will be high and the concentration of the reagents will be low. That's why the equilibrium constant has a likely magnitude of k>10³


Have a nice day!


pstnonsonjoku

The reaction tends to completion hence K should be in the order of K>103.

The equilibrium constant is a number that indicates the extent to which reactants are converted to products in a chemical reaction. A high value of equilibrium constant indicates that the system contains mostly products and few reactants at equilibrium. A low equilibrium constant indicates that the concentration of reactants is greater than the concentration of products at equilibrium.

For the reaction;  2H2(g) + O2(g) ⇌ 2H2O(g), we know the reaction to be explosive and tend to completion since it is a combustion reaction. This means that the reactants are mostly converted to products and the equilibrium constant will be large. Hence K should be in the order of K>103.

Learn more: https://brainly.com/question/10038290

Hydrogen is used as a rocket fuel because it is very light and reacts explosively and completely with oxygen. For the combustion reaction 2H2(g)+O2(g)⇌2H2O(g) what is the likely magnitude of the equilibrium constant K?

1. K<10−3

2. 10−3

3. K=0

4. K>103

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