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?

The equilibrium system N2O4(g) 2NO2(g) was established in a 1.00-liter vessel. Upon analysis, the following information was found: [NO2] = 0.500 M; [N2O4] = 0.0250 M. What is the value of Keq?

3.Which is the correct expression for Keq when the chemical reaction is as follows:
N2 + 3H2 2NH3.

4.Reactions which do not continue to completion are called ______reactions.

Answer 1

For reaction ax + by <=> cz + dm, the calculation of Keq is
`( [z]^(c) [m]^(d) )/([x]^(a) [y]^(b) )`. So the first Keq is 938450. The Second Keq is 10. The third expression is [NH3]2/([N2][H2]3). The last one is reversible.

Answer 2

Answer: 1)
`2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)`

Equilibrium constant is defined as the ratio of the product of concentration of products to the product of concentration of reactants each term raised to their stochiometric coefficients.

`K_(eq)=([H_2S]^2)/(H_2]^2* [S_2])`

where [] = concentration in Molarity=
`\frac{moles}{\text {Volume in L}}`

Thus
`[H_2S]=(68.5)/(1.0)=68.5M`

`[H_2]=(0.50)/(1.0)=0.50M`

`[S_2]=(0.020)/(1.0)=0.020M`

`K_(eq)=([68.5]^2)/(0.50]^2* [0.020])=938450`

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

2)
`N_2O_4(g)\rightleftharpoons 2NO_2(g)`

`K_(eq)=([NO_2]^2)/([N_2O_4])`

`K_(eq)=([0.500]^2)/([0.0250])=10`

3)
`N_2+3H_2\rightleftharpoons 2NH_3`

`K_(eq)=([NH_3]^2)/([N_2]* [H_2]^3)`

4) Reactions which do not continue to completion are called equilibrium reactions as the rate of forward reaction is equal to the rate of backward direction.