Question

at a certain temperature, the equilibrium constant ( k ) for the following reaction is ( 7.7\times10^{12} ):

( mathrm{co}(mathrm{g})+mathrm{h}_{2} mathrm{o}(mathrm{g})
ightleftharpoonsmathrm{co}_{2}(mathrm{g})+mathrm{h}_{2}(mathrm{g}) )

use this information to complete the following table.

suppose a 22. l reaction vessel is filled with 0.93 mol of ( mathrm{co}_{2} ) and 0.93 mol of ( mathrm{h}_{2} ). what can you say about the composition of the mixture in the vessel at equilibrium?

what is the equilibrium constant for the following reaction?
round your answer to 2 significant digits.

( mathrm{co}_{2}(mathrm{g})+mathrm{h}_{2}(mathrm{g})
ightleftharpoonsmathrm{co}(mathrm{g})+mathrm{h}_{2} mathrm{o}(mathrm{g}) )

what is the equilibrium constant for the following reaction?
round your answer to 2 significant digits.

( 2mathrm{co}(mathrm{g})+2mathrm{h}_{2} mathrm{o}(mathrm{g})
ightleftharpoons2mathrm{co}_{2}(mathrm{g})+2mathrm{h}_{2}(mathrm{g}) )

there will be very little ( mathrm{co} ) and ( mathrm{h}_{2} mathrm{o} ).

there will be very little ( mathrm{co}_{2} ) and ( mathrm{h}_{2} ).

neither of the above is true.

( k=square )

( k=square )

Explanation:

Step1: Analyze the first part

Given \(K = 7.7\times10^{12}\) for \(CO(g)+H_2O(g)
ightleftharpoons CO_2(g)+H_2(g)\). A large \(K\) value means the forward reaction is highly favored. But when we start with \(CO_2\) and \(H_2\), the reverse reaction will occur. Since \(K_{reverse}=\frac{1}{K_{forward}}=\frac{1}{7.7\times 10^{12}}\approx1.3\times10^{-13}\) (very small), the reverse reaction (formation of \(CO\) and \(H_2O\)) is not favored. So there will be very little \(CO\) and \(H_2O\).

Step2: Calculate \(K\) for \(CO_2(g)+H_2(g)

ightleftharpoons CO(g)+H_2O(g)\)
If \(K_1\) is for \(CO(g)+H_2O(g)
ightleftharpoons CO_2(g)+H_2(g)\) (\(K_1 = 7.7\times10^{12}\)), then for the reverse reaction \(CO_2(g)+H_2(g)
ightleftharpoons CO(g)+H_2O(g)\), \(K=\frac{1}{K_1}\).

$$K=\frac{1}{7.7\times 10^{12}}\approx1.3\times 10^{-13}$$

Step3: Calculate \(K\) for \(2CO(g)+2H_2O(g)

ightleftharpoons 2CO_2(g)+2H_2(g)\)
If \(K_1\) is for \(CO(g)+H_2O(g)
ightleftharpoons CO_2(g)+H_2(g)\) (\(K_1 = 7.7\times10^{12}\)), for \(nA + nB
ightleftharpoons nC + nD\), \(K=(K_1)^n\). Here \(n = 2\), so \(K=(7.7\times 10^{12})^2=5.9\times10^{25}\)

Answer:

• First part: There will be very little \(CO\) and \(H_2O\)
• Second part: \(K = 1.3\times 10^{-13}\)
• Third part: \(K = 5.9\times 10^{25}\)