When conducting chemical reactions in the lab or in industrial processes, it can

Question

When conducting chemical reactions in the lab or in industrial processes, it can be important to know whether a reaction has reached equilibrium. By measuring the reaction quotient, Q, of a chemical reaction and comparing it to the equilibrium constant, K, we can identify whether a reaction is at equilibrium.
For the reaction
aA+bBcC+dD
the reaction quotient, Q, is given by the expression
Q=[C]c[D]d[A]a[B]b
under any conditions. The value of Q is equal to K only at equilibrium.

2.

Part B

The following reaction was carried out in a 3.25 L reaction vessel at 1100 K:

C(s)+H2O(g)CO(g)+H2(g)

If during the course of the reaction, the vessel is found to contain 8.00 mol of C, 14.4 mol of H2O, 3.90 mol of CO, and 6.10 mol of H2, what is the reaction quotient Q?

Part C

The reaction

2CH4(g)C2H2(g)+3H2(g)

has an equilibrium constant of K = 0.154.

If 6.60 mol of CH4, 4.10 mol of C2H2, and 11.50 mol of H2 are added to a reaction vessel with a volume of 5.80 L , what net reaction will occur?

The reaction

has an equilibrium constant of  = 0.154.

If 6.60  of , 4.10  of , and 11.50  of  are added to a reaction vessel with a volume of 5.80  , what net reaction will occur?

No further reaction will occur because the reaction is at equilibrium.

3. At 700 K, Kp = 0.140 for the reaction ClF3(g)ClF(g)+F2(g).

Calculate the equilibrium partial pressures of ClF3, ClF, and F2 if only ClF3 is present initially, at a partial pressure of 1.59 atm .

The reaction will proceed to the left to establish equilibrium. The reaction will proceed to the right to establish equilibrium.

No further reaction will occur because the reaction is at equilibrium.

3. At 700 K, Kp = 0.140 for the reaction ClF3(g)ClF(g)+F2(g).

Calculate the equilibrium partial pressures of ClF3, ClF, and F2 if only ClF3 is present initially, at a partial pressure of 1.59 atm .

Explanation / Answer

Solution :-

Part B

Lets calculate the initial concnetrations of the each species using the given data

molarity = moles / volume in liter

[H2O] =14.4 mol / 3.25 L = 4.430 M

[CO] =3.90 mol / 3.25 L = 1.20 M

[H2]= 6.10 mol / 3.25 L = 1.877 M

now lets calculate the Qc

Qc= [CO][H2]/[H2O]

C is not involved in the Qc equation because it is in solid state

Qc= [1.20][1.877]/[4.430]

Qc =0.508

Therefore Qc= 0.508

So the reaction quotient Q = 0.508

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