at a certain temperature, the equilibrium constant $k$ for the following reaction is 574.:\n$co(g)+h_2o(g)\ri…

at a certain temperature, the equilibrium constant $k$ for the following reaction is 574.:\n$co(g)+h_2o(g)\rightleftharpoons co_2(g)+h_2(g)$\nuse this information to complete the following table.\nsuppose a 11. l reaction vessel is filled with 1.3 mol of co and 1.3 mol of $h_2o$. what can you say about the composition of the mixture in the vessel at equilibrium?\nthere will be very little co and $h_2o$.\nthere will be very little $co_2$ and $h_2$.\nneither of the above is true.\nwhat is the equilibrium constant for the following reaction? be sure your answer has the correct number of significant digits.\n$co_2(g)+h_2(g)\rightleftharpoons co(g)+h_2o(g)$\n$k = $\nwhat is the equilibrium constant for the following reaction? be sure your answer has the correct number of significant digits.\n$3co(g)+3h_2o(g)\rightleftharpoons 3co_2(g)+3h_2(g)$\n$k = $

at a certain temperature, the equilibrium constant $k$ for the following reaction is 574.:\n$co(g)+h_2o(g)\rightleftharpoons co_2(g)+h_2(g)$\nuse this information to complete the following table.\nsuppose a 11. l reaction vessel is filled with 1.3 mol of co and 1.3 mol of $h_2o$. what can you say about the composition of the mixture in the vessel at equilibrium?\nthere will be very little co and $h_2o$.\nthere will be very little $co_2$ and $h_2$.\nneither of the above is true.\nwhat is the equilibrium constant for the following reaction? be sure your answer has the correct number of significant digits.\n$co_2(g)+h_2(g)\rightleftharpoons co(g)+h_2o(g)$\n$k = $\nwhat is the equilibrium constant for the following reaction? be sure your answer has the correct number of significant digits.\n$3co(g)+3h_2o(g)\rightleftharpoons 3co_2(g)+3h_2(g)$\n$k = $

Answer

Explanation:

Step 1: Analyze equilibrium composition

A large ( K = 574 ) indicates the forward reaction is favored, so reactants ( \text{CO} ) and ( \text{H}_2\text{O} ) are mostly converted to products. Thus, there will be very little ( \text{CO} ) and ( \text{H}_2\text{O} ).

Step 2: Calculate reverse reaction equilibrium constant

The reverse reaction is the original reaction reversed, so its equilibrium constant is the reciprocal of ( K ):
( K' = \frac{1}{K} = \frac{1}{574} ).

Step 3: Calculate equilibrium constant for scaled reaction

When the reaction is multiplied by 3, the equilibrium constant is raised to the 3rd power:
( K'' = K^3 = 574^3 ).

Answer:

There will be very little CO and H₂O.
( K = \frac{1}{574} )
( K = 574^3 )