At a certain temperature, the equilibrium constant is 174 for
H2(g) + I2(g) ↔ 2HI(g)
Calculate the equilbrium constant for
0.5H2(g) + 0.5I2(g) ↔ HI(g)
For the same equation but with 1/2 the coefficients, the new eq is sqrt (174).
To calculate the equilibrium constant for the given reaction:
0.5H2(g) + 0.5I2(g) ↔ HI(g)
We need to use the relationship between the coefficients and the equilibrium constant. The equilibrium constant expression is given by:
K_eq = [HI]^2 / ([H2]^0.5 [I2]^0.5)
Where [A] represents the concentration of species A.
Since the coefficients of H2 and I2 are both 0.5, we can rewrite the equilibrium constant expression as:
K_eq = [HI]^2 / ([H2][I2])
Now, to calculate the equilibrium constant K_eq, we can use the given K_eq value for the previous reaction:
K_eq = 174 for H2(g) + I2(g) ↔ 2HI(g)
Substituting the known values:
174 = [HI]^2 / ([H2][I2])
To find K_eq for the new reaction, we need to determine the relationship between the concentrations in the two reactions.
For the new reaction: 0.5H2(g) + 0.5I2(g) ↔ HI(g)
The concentrations of the species are related by:
[0.5H2] = [H2]^0.5
[0.5I2] = [I2]^0.5
Substituting these relationships into the equation for K_eq:
174 = [HI]^2 / ([H2]^0.5 [I2]^0.5)
174 = [HI]^2 / ([0.5H2] [0.5I2])
174 = [HI]^2 / ([H2]^0.5 [I2]^0.5)
Therefore, the equilibrium constant for the reaction 0.5H2(g) + 0.5I2(g) ↔ HI(g) is also 174.
To calculate the equilibrium constant for the given reaction, we need to use the law of mass action. The law of mass action states that the equilibrium constant (K) is equal to the ratio of the product of the molar concentrations of the products, raised to the power of their stoichiometric coefficients, divided by the product of the molar concentrations of the reactants, also raised to the power of their stoichiometric coefficients.
In this case, the equilibrium constant for the first reaction, H2(g) + I2(g) ↔ 2HI(g), is given as 174. Therefore, we can write the expression for the equilibrium constant as:
K1 = [HI]^2 / ([H2] * [I2])
Now, let's calculate the equilibrium constant for the second reaction, 0.5H2(g) + 0.5I2(g) ↔ HI(g). Since the stoichiometric coefficients are different in this reaction, we need to adjust the equilibrium constant accordingly.
Given that the reaction is multiplied by 0.5 for both H2 and I2, we can rewrite the equation as:
H2(g) + I2(g) ↔ 2HI(g) (multiplied by 2)
Now, the molar concentrations for the second reaction can be related to those of the first reaction as:
[H2] = 0.5[H2] (from the second reaction)
[I2] = 0.5[I2] (from the second reaction)
[HI] = [HI] (because it is the same in both reactions)
Substituting these values into the expression for the equilibrium constant for the first reaction, we get:
K1 = [HI]^2 / ([H2] * [I2])
Substituting the adjusted concentrations for the second reaction:
K2 = [HI]^2 / (0.5[H2] * 0.5[I2])
Simplifying the equation:
K2 = 4 * [HI]^2 / ([H2] * [I2])
Therefore, the equilibrium constant for the second reaction is four times the equilibrium constant of the first reaction:
K2 = 4 * K1
Substituting the given value of K1 as 174:
K2 = 4 * 174
K2 = 696
Hence, the equilibrium constant for the reaction 0.5H2(g) + 0.5I2(g) ↔ HI(g) is 696.