What is the value for K_c for the following reaction: PbCl_2(s) Pb^2+(aq) + 2 Cl

Question

What is the value for K_c for the following reaction: PbCl_2(s) Pb^2+(aq) + 2 Cl^- (aq), if PbCl_2(s) = 1.50 grams, [Pb^2+] = 1.6 times 10^-2 M and [Cl^-] = 32 times 10^-2 M at equilibrium? (The molar mass of PbC1_2(s) is 278 g/mol and its density is 5.85 g/cm^3.) A) 16 times 10^-5 B) 7.6 times 10^-7 C) 1.3 times 10^6 D) 6.2 times 10^4 If K_c = 20 times 10^33 at 25 degree C for the following reaction: H_2(g) + C1_2(g) 2 HCl(g), then find K_p at the same temperature. A) 4.9 times 10^34 B) 8.2 times 10^31 C) 2.0 times 10^3 D) 9.7 times 10^32 An equilibrium mixture of CO, O_2 and CO_2 at a certain temperature contains 0.0010 M CO_2 and 0.0015 M O_2. At this temperature, K_o equals 1.4 times 10^2 for the reaction: 2 CO(g) + O_2(g) 2 CO_2(g). What is the equilibrium concentration of CO? A) 4.8 times 10^-6 M B) 3.1 times 10^-1 M C) 9.3 times 10^-2 M D) 2.2 times 10^-3 M Salt solubilities can be compared by the concentration of cation formed when the salt dissolves in the general reaction: M_a X_b (s) a M^b+(aq) + b X^a- (aq). Give the following salts and their equilibrium constants for the reaction above at 25 degree C, which salt is the least soluble? A) CaCO_3, K_c = 26 times 10^-9 B) AgCl, K_c 1.8 times 10^-10 C) Ag_2SO_4, K_c = 1.2 times 10^-5 D) CaF_2, K_c = 1.5 times 10^-10 For the reaction: 4 HCl(g) + 2(g) 2Cl_2(g) + 2 H_2O (l), the equilibrium constant is 0.063 at 400 K. If the reaction quotient is 0.100, which of the following statements is not correct? A) [HCI] will increase. B) [H_2O] will increase. C) [Cl_2] will decrease. D) [O_2] will increase.

Explanation / Answer

Q3 Solution : -

PbCl2 ---- > Pb^2+ + 2Cl^-

[Pb^2+] = 1.6*10^-2 M

[Cl-] = 3.2*10^-2 M

Kc = [Pb^2+] [Cl^-]^2

Kc = [1.6*10^-2] [3.2*10^-2]^2

Kc=1.6*10^-5

So the correct answer is option A

Q4 solution : -

H2+Cl2 --- > 2HCl

Kc = 2.0*10^33

Kp = Kc x (RT)^delta n

Delta n = 2- (1+1) = 0

Kp = 2.0*10^33 x (0.08206 L atm per mol K * 298 K)^0

Kp = 2.0*10^33

So the correct answer is option C

Q5 solution : -

2CO + O2 ---- > 2CO2

[CO2]= 0.0010 M , [O2] =0.0015 M

Kc= 1.4*10^2

[CO] equilibrium = ?

Consider the reverse reaction

2CO + O2 ---- > 2CO2

0             0.0015     0.0010

+2x             +x               -2x

+2x       0.0015+x      0.0010-2x

Kc = [CO2]^2 / [CO]^2[O2]

1/1.4*10^2 = ([2x]^2[0.0015+x] )/[0.001-2x]^2

Solving for x using quadratic equation we get

X= -0.00033 M

Equilibrium concentration of [CO]= 2x = (0.00033*2) = 0.00066 M

Not matching with given options.

Q6 solution :-

Lower the kc means less soluble is the salt therefore CaF2 has the lowest Kc therefore CaF2 will be the lest soluble salt hence answer is option D

Q7 solution :-

Kc= 0.063 and Qc = 0.100

When the Qc is greater than Kc then reverse reaction favors

Therefore the incorrect statement is H2O will increase therefore correct answer is option B

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