Question

For the reactions system 2H2(g) + S2(g) 2H2S(g), a 1.00 liter vessel is found to contain 0.50 moles of H2, 0.020 moles of S2, and 68.5 moles of H2S. Calculate the numerical value of the Keq of this system.

K =
Are the products or reactants favored?

Answer 1

Answer:

K = 9.4 *10^5

The reaction favor product formation

Explanation:

Equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium. Also it is defined as the ratio of the product of concentration of products to the product of concentration of reactants each term raised to power equal their stochiometric coefficients.  

so, for the reaction,

• 2H₂ ₍g₎ + S₂ ₍g₎  → 2H₂S ₍g₎

The equilibrium constant can be expressed as following:

$K_{eq}=\frac{[H_2S]^2}{H_2]^2\times [S_2]}$

where  concentration of reactants and products is expressed in molarity Molarity=(no of moles/ Volume L)

So,

[H₂] = (0.50 mol / 1 L) = 0.50 M

[S₂] = (0.02 mol / 1 L) = 0.020 M

[H₂S] = (68.5 mol / 1 L) = 68.5 M

∴ K_{eq}=\frac{[68.5]^2}{0.50]^2\times [0.020]}=9.4*10^5

As the value of K is greater than 1, the reaction favor product formation.

Answer 2

Answer:

B, I Took the test

Explanation:

For the reactions system 2H 2 (g) + S 2 (g)   2H 2 S(g), a 1.00-liter vessel is found to contain 0.50 mole of H 2 , 0.020 mole of S 2 , and 68.5 moles of H 2 S. Calculate the numerical value of the K eq of this system.

K = _____

A. 6.9 x 10 3

B. 9.4 x 10 5

C. 1.1 x 10 6

D. 1.4 x 10 4