Question

Consider the equation: 2NO(0) - N.04(). Using ONLY the information given by the equation which of the following

Answer 1

Answer:

By increasing the pressure, the molar concentration of  N2O4 will increase

Explanation:

We have the equation 2NO2 ⇔ N2O4

This equation is reversible and exotherm. By decreasing the temperature, the reaction will produce more energy, so the reaction will move to the right.  But a lower temperature also lowers the rate of the process, so, the temperature is set at a compromise value that allows N2O4 to be made at a reasonable rate with an equilibrium concentration that is not too unfavorable

So increasing the temperature will shift the equilibrium to the left. The equilibrium shifts in the direction that consumes energy.

If we decrease the concentration of NO2, the equilibrium will shift to the left, resulting in forming more reactants.

To increase the molar concentration of the product N2O4, we have to increase the pressure of the system.

NO2 takes up more space than N2O4, so increasing the pressure would allow the reactant to collide more form more product.

By increasing the pressure, the molar concentration of  N2O4 will increase

Answer 2

Answer:

5) Increase the pressure

Explanation:

$2NO_2 (g)N_2 O_4 (g) \\\\\Delta H=-58 KJ per mol$

In a chemical reaction, equilibrium is the state in which the rate of the forward reaction is equal to the rate of the reverse reaction.  A system will remain in equilibrium unless it is stressed or disturbed.  Le Chatelier’s Principle states that “when a stress is placed on a system at equilibrium, the system will shift to offset the stress applied”.

Equilibrium always shifts away from the increase and towards the decrease.

The equation here shows us that forward reaction is exothermic since ∆H is negative and backward or reverse reaction is endothermic.

Increasing the temperature will shift the equilibrium in favour of the endothermic reaction.

Decreasing the temperature will shift the equilibrium in favour of exothermic reaction.

Increasing the Pressure towards the side with lesser number of gaseous moles

Decreasing the Pressure towards the  side with more number of gaseous moles.

Increasing the concentration of the substance favour the equilibrium shift away from the substance

Decreasing the concentration of the substance favour the  equilibrium shift towards the substance.

$2NO_2 (g)N_2 O_4 (g)$ $\Delta H$ is not given

Since delta H is not given we can rule out options 1 and 3. Decreasing the concentration of NO favours equilibrium shift towards the left side so $N_2 O_4$ is not produced in greater amount.

So, taking into pressure conditions,

Left side contains 2 moles and right side contains 1 mole.

Increasing the Pressure will shift the equilibrium towards the lesser number of moles that is right side producing more of $N_2 O_4$.