1. The reaction below has an equilibrium constant Kp=2.2×106 at 298 K. 2COF2(g)C

Question

1. The reaction below has an equilibrium constant Kp=2.2×106 at 298 K. 2COF2(g)CO2(g)+CF4(g)

Calculate Kp for the reaction below.
COF2(g)12CO2(g)+12CF4(g)

Calculate Kp for the reaction below.
23COF2(g)13CO2(g)+13CF4(g)

Calculate Kp for the reaction below.
2CO2(g)+2CF4(g)4COF2(g)

2. For the reactionA(g)2B(g), a reaction vessel initially contains only A at a pressure of PA=1.24 atm . At equilibrium, PA =0.19 atm . What is Kp

3. For the reaction 2A(g)B(g)+2C(g), a reaction vessel initially contains only A at a pressure of PA=255 mmHg . At equilibrium, PA=70 mmHg . What is Kp.

4. Consider the following reaction:
Fe3+(aq)+SCN(aq)FeSCN2+(aq)
A solution is made containing an initial [Fe3+] of 1.1×103 M and an initial [SCN] of 7.8×104 M. At equilibrium, [FeSCN2+]= 1.8×104 M .

Calculate the value of the equilibrium constant (Kc).

Explanation / Answer

1. When,

2COF2(g) <==> CO2(g) + CF4(g) .... Kp = 2.2 x 10^6

So,

2/3COF2(g) <==> 1/3CO2(g) + 1/3CF4(g) ... Kp = cube rt.(2.2 x 10^6) = 130.059

For,

2CO2(g) + 2CF4(g) <==> 2COF2(g) .... Kp = 1/(2.2 x 10^6)^2 = 2.066 x 10^-13

2. Kp = [B]^2/[A] = (2 x (1.24 - 0.19))^2/(0.19) = 23.21

3. change in PA = 255 - 70 = 185 mmHg

Kp = [B][C]^2/[A]^2 = (185)(2 x 185)^2/(70)^2 = 1292.17

4. [Fe3+]eq = 1.1 x 10^-3 - 1.8 x 10^-4 = 9.2 x 10^-4 M

[SCN-]eq = 7.8 x 10^-4 - 1.8 x 10^-4 = 6 x 10^-4 M

Kc = [FeSCN]-eq/[Fe3+]eq[SCN-]eq = (1.8 x 10^-4)/(9.2 x 10^-4) x (6 x 10^-4) = 326.087

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