How would increasing the H_2 (g) concentration affect the following equilibrium?

Question

How would increasing the H_2 (g) concentration affect the following equilibrium? N_2 (g) + 3 H_2 (g) 2 NH_3(g) Under certain condition, hydrogen gas will combine with oxygen gas to produce gaseous water. The following equilibrium results. The forward reaction is exothermic. 2H_2 (g) + O_2 (g) 2 H_2 O(g) How would an increase in temperature affect the equilibrium? At equilibrium, the following concentrations exist: [H_2] = 1.14 Times 10^-2 M; [I_2] = 1.20 Times 10^-3 M; and [HI] = 2.52 Times 10^-2 M. Calculate the equilibrium constant for this reaction. H_2 (g) + I_2 (g) rightarrow 2Hl (g) Given that equilibrium constant expression, write the appropriate chemical reaction. K = [HCl]^4 [O_2]/[Cl_2]^2 [H_2 O]^2

Explanation / Answer

we know that

according to Le chatlier principle

the equilibrium will shift in a direction to counter the change

13)

the reaction is

N2 + 3H2 ---> 2 NH3

given

H2 conc is increased

so

the equilibrium will shift in a direction to decrease the conc of H2

so

it will shift to the right

14)

2H2 + 02 --> 2H20

given exothermic reaction

so the reaction can be written as

2H2 + 02 --> 2H20 + heat

given

increase in temperature ,that is heat is added

so

the equilibrium will shift in a direction to reduce the heat

as a result

the equilibrium will shift to the left

15)

given

H2 + i2 ---> 2 HI

the equilibrium constant is given by

K = [HI]^2 / [H2] [I2]

using given values

K = [ 2.52 x 10-2 ]^2 / [ 1.14 x 10-2 ] [1.2 x 10-3]

K = 46.4

so

the equilibrium contant is 46.4

16 )

given equilibrium constant is

K = [HCl]^4 [02] / [ Cl2]^2 [ H20]^2

numerator are products

and

denominator are reactants

and the reaction is given by

2 Cl2 + 2 H20 --> 4 HCl + 02

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