Prof. M 2. In the U.S. natural gas (methane) is obtained from deep wells. Howeve

Question

Prof. M 2. In the U.S. natural gas (methane) is obtained from deep wells. However, in countries that do not have wells for natural gas, a replacement is a mixture of carbon monoxide and hydrogen. This is called water gas. Both gases are able to burn and the mixture can be used in the same way that we use natural gas. The chemical reaction for making water gas is C(s) + H2O(g) CO(g) + H2(g) Into a 100.0 L vessel maintained at 900 K was placed 40.0 g of solid carbon, 4.00 atm of water, 0.200 atm of hydrogen and 0.700 atm of carbon monoxide and these were allowed to reach equilibrium, at which time the partial pressure of water was 3.22 atm and there were 35.8 g of carbon. The partial pressure of carbon monoxide was 0.975 and that of hydrogen was 1.48 atm. a. In the box on the right, give the equilibrium constant b. The value of the reaction quotient in terms of pressure before the reaction began was c The value of the equilibrium constant in terms of pressure for this reaction is expression in terms of pressures for the reaction above. d. T he addition of hydrogen to the vessel at equilibrium will produce (circle all that apply): more carbon more water more carbon monoxide no change insufficient information e. The addition of carbon to the vessel, when contents at equilibrium, will produce (circle all that apply): more hydrogen more : The draw the Lewis dot structure for carbon monoxide in the box ONUS Assuming that the volume occupied by the carbon negligible, how many grams of hydrogen were initially the vessel. ans water more carbon monoxide no change insufficient information CO

Explanation / Answer

Q2

a

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Kp expression:

must be based on pressures

Kp = (P-CO * P-H2)/(P-H2O)

b)

For Q:

Qp = (P-CO * P-H2)/(P-H2O) (species NOT in equilbirium)

Qp = (0.7*0.2)/(4) = 0.035

c)

Kp in equilbirum

Kp = (0.975*1.48)/(3.22)

Kp = 0.448

d)

addition of H2, a product, will make a left shift, so CO decreases, H2O increases

also, more cabron will be formed

e)

addition of C(s) --> has no effect, since it is a solid

f)

C(triple)O

valence C = 4, vlaence O = 6

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