Coal can be used to generate hydrogen gas (a potential fuel) by the following endothermic reaction:C(s)+H2O(g) >CO(g) + H2(g)If this reaction mixture is at equilibrium, predict whether each of the following will result in the formation of additional hydrogen gas, the formation of less hydrogen gas, or have no effect on the quantity of hydrogen gas:.A. adding more C to the reaction mixtureB. adding more H2O to the reaction mixtureC. raising the temperature of the reaction mixtureD. increasing the volume of the reaction mixtureE. adding a catalyst to the reaction mixtureF. adding an inert gas to the reaction mixture

A. No effect

B. Results in the formation of additional hydrogen gas

C. Results in the formation of additional hydrogen gas

D. Results in the formation of additional hydrogen gas

E. No effect

F. No effect

Explanation:

The equilibrium in this question is

C(s) + H₂O (g) ⇄ CO(g) + H₂ (g)

and

Kp = pCO x pH₂/ pH₂O

where pCO, pH₂O and pH₂O are the partial pressures of CO, H₂ and H₂O.

We call the equilibrium constant Kp since only gases intervene in the expression for the constant.

A. adding more C to the reaction mixure

Adding more carbon which is a solid does not alter the  pressure equilibrium constant, therefore, it has no effect on the equilibrium and consequently no effect on the quantity of hydrogen gas.

B. adding more H₂O to the reaction mixture

We can answer this part by using  Le Chatelier's principle which states that a system at equilibrium will respond to a stress in such a way as to minimize the stress, hence  restoring equilbrium.

One of the three possible stresses is an increase of reactant as in this case. The system will react by decreasing some of the added water. Thus the equilbrium shifts to the product side which will result in the formation of more hydrogen gas.

The difference of this part with respect to part A is that indeed the water gas is included in the equilibrium constant expression.

C. raising the temperature

This is another stress we can subject an equilibrium.

We are told the reaction is endothermic which means in going from left to right it consumes heat. Thus the equilibrium will shift to the product side by consuming some of the added heat favoring the production of more hydrogen gas.

D. increasing the volume of the reaction mixture

This the last of the stresses .

Increasing the volume of the reaction effectively decreases the pressure ( volume is inversely proportional to pressure ) so the equilibrium will shift to the side that has more pressure which is the product side: we have two moles of gases  products  vs. 1 mol gas in the reactant side.

Therefore, the equilibrium will shift to the right increasing the quantity of H₂.

E. adding a catalyst to the reaction mixture

The addition of a catalyst does not have an effect on the equilibrium constant. The catalyst will speed both the forward and reverse reaction decreasin the time to attain equilibrium.

So there is no effect on the quantity of H₂.

F. Adding an inert gas to reaction mixture

Assuming the volume of the reaction mixture remains constant, and we are not told such change in volume occurred, the addition of an inert gas does not have an effect in our equilibrium. The inert gasdoes not participate  in the calculation for Kp.

The situation will be different if the volume of the reaction is allowed to increase, but again this is not stated in the question.

answer; /// i do believe that the correct answer is (c)nuclear bond