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.

For reaction ax + by <=> cz + dm, the calculation of Keq is \(\frac{ [z]^{c} [m]^{d} }{[x]^{a} [y]^{b} }\). 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.

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kobenhavn kobenhavn · Answer 2 · 2019-01-24T02:46:35-05:00

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.

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

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

Thus \([H_2S]=\frac{68.5}{1.0}=68.5M\)

\([H_2]=\frac{0.50}{1.0}=0.50M\)

\([S_2]=\frac{0.020}{1.0}=0.020M\)

\(K_{eq}=\frac{[68.5]^2}{0.50]^2\times [0.020]}=938450\)

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

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

\(K_{eq}=\frac{[NO_2]^2}{[N_2O_4]}\)

\(K_{eq}=\frac{[0.500]^2}{[0.0250]}=10\)

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

\(K_{eq}=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}\)

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.