Section Test:Chemical Equilibrium

Section Test:Chemical Equilibrium

From MyMCAT

Jump to: navigation, search

Topics Covered

This Section Test deals with the concepts of chemical equilibria and Le Chatelier's Principle which describes the effects of changes to equilibrium.

Section Test

1. What is the relationship between a chemical equilibrium and the rates of its forward and reverse reaction?

In equilibrium, the forward reaction rate must be greater than the reverse reaction rate.
In equilibrium, the forward reaction rate must be less than the reverse reaction rate.
In equilibrium, the forward and reverse reaction rates must be equal.
In equilibrium, both forward and reverse reactions must stop.

2. The appropriate expression for the equilibrium constant Kp for the gas phase reaction between nitrogen and hydrogen to form ammonia is?

(PH2)3(PN2)/(PNH3)2
(PNH3)2/(PH2)3(PN2)
PNH3/(PH2)(PN2)
(PH2)(PN2)/PNH3
→ To write the Keq (or Kp) expression in this case, we need to determine what the reaction is doing. Here we have H2 and N2 forming NH3. To make this reaction complete, we need to balance the moles of each reactant and product. The only way to do this is to use 3 moles of H2 and 1 of N2 to make 2 moles of NH3. Or, 3 H2 + 1 N2 ⇌ 2 NH3. The Kp expression is then the partial pressures of each product to the power of the coefficient over the reactants to the power of their coefficient (or the answer C).

3. For the equilibrium N2O3(g) ⇌ NO(g) + NO2(g), which of the following is correct?

Increasing the [NO] while shift the equilibrium towards products.
Increasing the pressure by decreasing the volume will shift the equilibrium left.
Increasing the [N2O3] will shift the equilibrium to the right.
Adding a catalyst will shift the equilibrium to the products.
→ Increasing the pressure will result in the system trying to relieve this increase. Because the left/reactants side has 1 mole equivalents while the right/products side has 2 mole equivalents, an increase in the pressure will shift to the side with fewer moles, or to the left/reactants.

4. For the equilibrium 2 SO2 (g) + O2 (g) <--> 2 SO3 (g) + heat, ___________________ will shift the system to the right. { For the equilibrium N2O3(g) ⇌ NO(g) + NO2(g), which of the following is correct?

a decrease in O2
an increase in volume
an increase in temperature
a decrease in SO3
→ When SO3 is removed from the system, the equilibrium must shift to the right to replenish this loss. All the other options result in a shift to the left. Note that an increase in volume will cause a decrease in pressure so the system will go to the side which regains some pressure, which is the left.

5. For the equilibrium N2 (g) + 3H2 (g) <--> 2NH3 (g) ,       Kp = 1 x 10-4, which of the following statements is false?

The rate constant for the forward reaction is less than that of the reverse reaction.
The value of the activation barrier cannot be determined from the data given.
The equilibrium constant is given by KC = [NH3]2 / [N2][H2]3
Conducting the reaction under low pressure will increase the yield of ammonia.
→ low pressure will result in a shift to the left to regain pressure in the system, this is contrary to trying to produce ammonia and is thus the wrong answer.

6. The the equilibrium CH4 (g) + 4Cl2 (g) <--> CCl4 (liq) + 4 HCl(g)      DH0 = -398 kJ/mol, which of the following statements is true?

The equilibrium is shifted to the left if the temperature is raised
The equilibrium is shifted to the right if the pressure is lowered
The equilibrium is unaffected if some carbon tetrachloride is removed
The equilibrium is shifted to the left if some hydrogen chloride is added
→ Pure liquids are never included in an equilibrium equation as their concentration to the rest of the elements is meaningless. Thus carbon tetrachloride removal makes no difference.

7. A mixture of 0.75 mol of N2 and 1.20 mol of H2 are placed in a 3.0 liter container. When the reaction, N2 (g) + 3H2 (g) <--> 2NH3 (g), reaches equilibrium, [H2] = 0.100 M. What is the value of [N2] and [NH3] at equilibrium?

[N2] = 0.150; [NH3] = 0.200
[N2] = 0.200; [NH3] = 0.150
[N2] = 0.300; [NH3] = 0.100
[N2] = 0.450; [NH3] = 0.050
→ To answer this problem, we need to understand that initially, the system is NOT at equilibrium because we have no products (only the N2 and H2) however after some time, the system will equalize itself to become an equilibrium.
→ Initially, we have 0.75 mol of N2 and 1.20 mol of H2. As time proceeds however, we will start to make the product 2NH3. If we were to assume that "x" moles of 2NH3 were made, then it would have taken "x" moles of N2 and "3x" moles of N2 to produce "2x" moles of NH3. We also know that the final concentration of H2 is 0.1M, which is 0.3 moles (since its a 3L volume). Thus H2 went from 1.2 moles down to 0.3 moles. This implies that the 3x change is 0.9, or x = 0.3. From this we can determine that the final N2 is 0.75-0.3 = 0.45 moles or 0.45moles/3L = 0.15M. For the H2 we have 1.5-3(0.3)=0.6 or 0.6moles/3L = 0.2M (This was given already in the question but we have done it here again to double check nothing was calculated wrong). and for the NH3 we went from 0 to 2(0.3) = 0.6 moles or 0.6moles/3L = 0.2M.
→ It is important to note that this problem did NOT require the use of the equilibrium expression. This was only possible because we were given the final concentration of one of the compounds. If we did not know this concentration, we would have had to use a Keq value and use the Keq expression to determine how all the compounds change.

8. 1.40 moles of H2 and 1.40 moles of I2 are placed in a 1.00 L vessel. What is the equilibrium concentration of HI in the gaseous mixture assuming the reaction is H2(g) + I 2(g) ⇌ 2HI(g) and Keq=49.0M-1?

2.18M
2.96M
4.36M
5.84M
→ First let us write out the Keq expression,
→K_{eq} = \frac{ \left[ HI \right]^{2} }{ \left[ H_{2} \right]\left[ I_{2} \right] } = 49.0
→ Now let us consider what happens as this system reaches equilibrium. H2 and I2 will each decrease in concentration as the product is formed. Thus if "2x" moles of product are made, we would need "x" moles of H2 and "x" moles of I2. If we assume this is what happens and plug it into the Kep equation we get,
→ K_{eq} = 49.0 = \frac{ \left[ HI \right]^{2} }{ \left[ H_{2} \right]\left[ I_{2} \right] } = \frac{ \left[ 2x \right]^{2} }{ \left[ H_{2} - x \right]\left[ I_{2} - x \right] }
→ Plugging in values, we then get,
→ 49.0 = \frac{ \left[ 2x \right]^{2} }{ \left[ H_{2} - x \right]\left[ I_{2} - x \right] } = \frac{ \left[ 2x \right]^{2} }{ \left[ 1.4 - x \right]\left[ 1.4 - x \right] }
→ Taking the squareroot of both sides can then simplify this complex equation,
→ 7 = \frac{ \left[ 2x \right] }{ \left[ 1.4 - x \right] } and then solving for x we get x=1.09M and thus 2x = [HI] = 2.18M.

9. At equilibrium, the reaction PCl3 (g) + Cl2 (g) <--> PCl5 (g) + 4 HCl(g) has a Keq of 96.2 M-1 at 400K. What is the concentration of PCl5 if, at 400K, the equilibrium concentrations of PCl3 (g) + Cl2 (g) are ).5M and 0.07M respectively?

[PCl5 (g)] = 0.425M
[PCl5 (g)] = 0.850M
[PCl5 (g)] = 1.70M
[PCl5 (g)] = 3.40M
→ We are given two of the three concentrations at equilibrium and the Keq. To solve for the third concentration then, we should determine the Keq expression and solve for the missing term.
→ K_{eq} = 96.2 = \frac{ \left[ PCl_{5}\right] } {\left[ PCl_{3} \right] \left[ Cl_{2} \right] }
→ rearranging and solving then gives us [PCl5 (g)] = (

10. For the equilibrium reaction, Fe2O3 (s) + 3 CO(g) --> 2Fe (s) + 3 CO2(g), which of the following expressions correctly inter-relates Kc and Kp?

Kc = Kp (RT)
Kp = Kc (RT)
Kc = Kp (RT)^2
Kp = Kc
→ Kc is equivalent to the Keq expression when we are referring to concentrations while Kp is equivalent to the Keq when we are refering to partial pressures. To relate these two, we should first consider some of the important concepts. In this reaction only CO and CO2 are important because the others are both solids and thus not included in the Keq expression. Furthermore, the number of moles of gases from reactants to products is three to three, implying there is NO change to the pressure when this reaction occurs. Now we must consider how concentration and pressure are related. The only formula we use for pressure calculations is PV = nRT. If we rearrange this for P, then for any gas in the system is P= (n/V)RT, but n/V is really the same thing as moles/liter, or M. So for a gas A, PA=[A]RT. If we now look at our Keq expression we have,
→K_{eq} = \frac{ \left[ CO_{2} \right]^{3} }{ \left[ CO \right]^{3} }
→ For Kp then, we can do the following,
→K_{p} = \frac{ P_{CO_{2}} }{ P_{CO} } = \frac{ \left[ CO_{2} \right]^{3}\left( RT \right)^{3} }{ \left[ CO \right]^{3}\left( RT \right)^{3} }
→Canceling out the like terms, we actually remove the (RT)3 terms altogether and we are left with Kc equals Kp. This is a special case for when the number of moles of gases does not change between the reactants and products, in any other system this is generally not the case and they will not be equal!

Your score is 0 / 0
Retrieved from "http://mymcat.com/wiki/Section_Test:Chemical_Equilibrium"