Consider the following system, which is at equilibrium,
3C(s) + 3 H2(g) mc020-1.jpg CH4(g) + C2H2(g)
The result of removing some C(s) from the system is that
a) more CH4(g) and C2H2(g) are produced to replace that which is removed.
b) Kc decreases.
c) more C(s) is produced.
d) no further change occurs.
The given system is in equilibrium, meaning that the forward and backward reactions are occurring at equal rates.
According to Le Chatelier's principle, when a change is made to a system at equilibrium, the system will respond in a way to counteract that change.
In this case, if some C(s) is removed from the system, the concentration of C(s) will decrease. To counteract this decrease, the system will shift in a direction that produces more C(s).
The reaction indicates that the forward reaction (3C(s) + 3H2(g) -> CH4(g) + C2H2(g)) produces C(s), so the system will shift to the right, producing more CH4(g) and C2H2(g) to replace the C(s) that was removed.
Therefore, option (a) is correct: more CH4(g) and C2H2(g) are produced to replace that which is removed.
To determine the result of removing some C(s) from the system, we need to consider the reaction stoichiometry and Le Chatelier's principle.
In the given equilibrium reaction:
3C(s) + 3H2(g) ⇌ CH4(g) + C2H2(g)
Removing some C(s) from the system will disrupt the equilibrium since the concentration of C(s) is changing. According to Le Chatelier's principle, the system will try to counteract this change by shifting the equilibrium in a direction that minimizes the effect of the disturbance.
In this case, since C(s) is a reactant, removing some C(s) will decrease the concentration of reactants and disrupt the balance of the equation. To restore equilibrium, the reaction will shift to the right, favoring the formation of products.
Therefore, the correct answer is:
a) more CH4(g) and C2H2(g) are produced to replace that which is removed.