chemistry 9950151 2
Initially, we place 2 mol of A and 4 mol of B in a 0.5 L flask. At equilibrium, the flask contains 0.3 mol of C. Determine the value of K.
We place 1 mol of A and 1.5 mol of B in a 250 mL container. We find that, at equilibrium, we have 0.5 mol of B. Calculate the value of K.
A system is represented by the following equation:
At equilibrium, [C]=0.6 M. Calculate the value of K if initially 1 mol of A and 2 mol of B were placed in a 1 L flask.
In a 500 mL flask, we put 6 mol of A, 6 mol of B, 4 moles of C and 9 mol of D. The value of K is 17.3.
Is the system in equilibrium? If not, in which direction will the reaction proceed to attain equilibrium?
A 100 mL container is filled with 0.01 mol of A2B3. At equilibrium, it also contains 0.006 mol of B. Find the value of K.
You are given the following equation:
You fill a 1 L flask with 0.20 mol of A2B. 0.5% of this substance dissociates. What is the value of K?
5.6 x 10-6 mol of A and 5 x 10-5 mol of B are mixed in a 200 mL flask. The system is represented by the equation:
At equilibrium, there is 4.8 x 10-5 mol of B. Calculate the value of the equilibrium constant.
K = 1.7 x 10-10 for the following system, which represents the dissolving of silver chloride:
What will the concentration of Ag+ be at equilibrium (in M)?
You work with the following system:
At, equilibrium, [Y]=3[X]. If K=10, calculate the [Y] at equilibrium.
- A chemical system within a sealed 1 L reaction vessel is described by the following reversible reaction equation:
2H2S(g) <–>2H2(g) + S2(g)
If the equilibrium constant is 0.000 004 200 at 1103 K find:
- the reaction quotient initially.
- the order of concentration of all three components at equilibrium (greatest concentration to lowest) without using calculations.
- what the size of the reaction quotient indicates regarding the extent of the forward reaction.
- the equilibrium concentration of sulphur gas if 0.070 mol of hydrogen sulphide gas is initially placed in the vessel.
Note: for part d you may make the simplifying assumption that the amount of hydrogen sulphide gas reacted is extremely small in relation to the initial amount of hydrogen sulphide gas—this will allow you to avoid a quadratic expression in the denominator of the reaction quotient by disregarding one term).