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20. If K =(0.92) at a (150°C) for the equilibrium reaction: CO(g) + 3H2(g) CH(g) +...

Question

20. If K =(0.92) at a (150°C) for the equilibrium reaction: CO(g) + 3H2(g) CH(g) +...

20. If K =(0.92) at a (150°C) for the equilibrium reaction: CO(g) + 3H2(g) CH(g) + H2O(g) The K of the reaction is : A) 5.63x
20. If K =(0.92) at a (150°C) for the equilibrium reaction: CO(g) + 3H2(g) CH(g) + H2O(g) The K of the reaction is : A) 5.63x 104 B) 2.32 x10+ C) 4.20 x105 D ) 3.29 x 109. Putor (*): 1. Reaction in which entire amount of the reactants is not converted into products is termed as reversible reaction ( ). 2. For exothermic reactions raising the temperature causes a shift to the right (X). 3. If the number of moles of product is reduced This type of reactions favor high pressure for product formation (). 4. law of mass action is state the rate of chemical reaction is directly proportional to concentration of reactant (V). 5. endothermic reactions favor low temperature for products formation (x ). 6. K = K (RT)An (*). 7. this reaction CH 6 + H06C0+3 HC) is Homogeneous equilibria () 8. reversible Reaction can proceed to completion(). 9. For the reaction CH (8) + H 0(g) = CO(g) + 3H2(g), favor low pressure for product formation(). 10. Kc for a heterogeneous equilibrium, you include a solid state in the equilibrium expression.

Answers

we know that (xp = E(Rajan An = number of moles of gaseous products mpl – number of moles of gaseous reactants any) an = np-n


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