Can someone check my answers for the following questions...btw HAPPY HOLIDAYS TO EVERY AT JISKHA!
1)As the magnitude of Keq increases, the amount of product formed in a reversible reaction: REMAINS CONSTANT
2)The Keq of reacion 4 x 10-7. At equilibrium: THE REACTANTS ARE FAVORED
3)To increase the yield of CH3OH for the equilibrium reaction CO(g) + 2H2 (g) <--> CH3OH (g) + 24 kcal: INCREASE THE T AND DECREASE THE P
1) Consider again the definition of Keq. The concentrations of products appear in the numerator, and the concentrations of reactancs appear in the numerator. If Keq increases, the concentrations of products must increase.
2) Correct
3) Apply Le Chatelier's principle here. The forward reaction is exothermic and it also decreases the number of moles in a fixed volume. To increase yield, you want to favor the forward reaction. Increasing temperature and decreasing pressure favors the REVERSE reaction, by tending to compensate for the temperature rise (absorbing heat) and pressure decrease (producing more moles)
1) To check the answer for the first question, you need to understand the concept of equilibrium and the expression for the equilibrium constant, Keq. Keq is a ratio of the concentrations (or partial pressures if dealing with gases) of the products to the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients.
In a reversible reaction, at equilibrium, the forward and backward reactions occur at the same rate, and the concentrations of reactants and products remain constant. If the magnitude of Keq increases, it means that the concentration of products compared to the reactants is higher. Therefore, as the magnitude of Keq increases, it indicates that the reaction favors the formation of products over reactants. So, the answer is INCORRECT, as the amount of product formed does not remain constant but rather increases as Keq increases.
2) In the second question, you are given the value of Keq as 4 x 10^(-7). Keq is calculated by taking the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients. A Keq value less than 1 indicates that the reactants are favored, meaning that the concentration of reactants is higher compared to the concentration of products at equilibrium.
Therefore, the answer is CORRECT. At equilibrium, with a Keq value of 4 x 10^(-7), the reactants are favored and their concentrations are higher.
3) The third question requires an understanding of how changes in temperature and pressure affect the yield of a reaction. The given equilibrium reaction is:
CO(g) + 2H2(g) <--> CH3OH(g) + 24 kcal
To increase the yield of CH3OH, you need to shift the equilibrium towards products by manipulating the temperature and pressure. Here are the correct steps:
- Increase the temperature (T): According to Le Chatelier's principle, if you increase the temperature, the equilibrium will shift in the endothermic direction. In this case, it means shifting right, favoring the products. Therefore, increasing the temperature will increase the yield of CH3OH.
- Decrease the pressure (P): If a reaction involves gases, decreasing the pressure will shift the equilibrium to the side with fewer moles of gas. In this case, by reducing the pressure, the equilibrium will shift towards the products (CH3OH), as it has only one mole of gas compared to three moles of gas in the reactant side (CO + 2H2). So, decreasing the pressure will increase the yield of CH3OH.
Therefore, the answer is CORRECT. To increase the yield of CH3OH, you need to increase the temperature (T) and decrease the pressure (P).