Question

Report an average equilibrium constant for the reaction at 25 degrees Celsius and the equilibrium constant for the reaction at 100 degrees Celsius. (A-D were done at 25 degrees Celsius, and E was done at 100 degrees Celsius)

ita Table Trial Initial Colar Final Color CO(Hab) &*] M ICH ()000.210 ko A red Ted 0.9605391.868 0.03346 12 0.064605 B red purple 0.296 733 4.43693 10.326267 0.249332 IC red bright led 0.769230 0.16384 6.60000 2247 2.459721 D purple magenta 01415791 2.39 044 0.03675.0.03.0990 E "Red grey 10.6419091 0.967634013550910.732764 Kc = [cool [contt20)6] Eco Trial A Cb. b3Buke 21 10 9665397 1.8666 - 10.064605 11 Trial B K= 0.328267) (10291973ac443693] -0.249332 Trial a [o.co.00122992) 2.4597€ - 7 20,769230100.1538413 Trial p KE D0.03975 +02 0.638990 0.41579132.390440 Thia K= [0, 358091 0.982769 50.641909] [0.567634

Answer 1

the problem is based on the concepts of the reaction with the transition metals.

when octahedral hexaaquacobalt(II) ion
Co(H20)6]2+
reacts with
CI
ion it will lead to formation of a blue colored solution. the reaction will be,

Co(H20)6]2+ (aq) + 4C1-(aq) + COC14]2-(aq) + 6H2O(1)

at the equilibrium , we can write the equilibrium constant as,

...................... 1

K [Coci) [C1-14 * Co(H20)2+1

so,

we have to find the average equilibrium constant for this reaction at 25 C and 100 C. the data of concentration is given to us as,

Trial (Co(H20)6]2+ (M) (Cl-) (M) (CoCl4]2-(M) A 0.966593 1.86616 0.0334612 B 0.296733 4.43693 0.328267 с 0.76923 0.153841 0.00000122 D 0.415791 2.39044 0.03875 E 0.641909 0.567634 0.358091

so we are given that trial A to D , occurs at 25 C. so K calculated for A to D will be for 25 C, hence,

we use the equation 1,

for trial A,

K [Coci) [C1-14 * Co(H20)2+1

we get

$K = \frac{[0.0334612 M]}{[1.86616M]^4 * [0.966593M]} = 0.00285 M^{-4}$

for trial B,

K [Coci) [C1-14 * Co(H20)2+1

we get

$K = \frac{[0.328267 M]}{[4.43693M]^4 * [0.29673M]} = 0.002854 M^{-4}$

for trial C,

K [Coci) [C1-14 * Co(H20)2+1

we get

$K = \frac{[0.00000122 M]}{[0.153841M]^4 * [0.76923M]} = 0.002831 M^{-4}$

for trial D,

K [Coci) [C1-14 * Co(H20)2+1

we get

$K = \frac{[0.03875 M]}{[2.39044M]^4 * [0.41579M]} = 0.002854 M^{-4}$

so, we have calculated K values for A to D trial , so now we take the average to get average equilibrium constant at 25 C.

we get

$K_{@25 C} = \frac{0.002854 M^{-4} + 0.002831 M^{-4} +0.002854 M^{-4}+0.002854 M^{-4}}{4}$

we get

...................[ans.]

hence for 25 C the equilibrium constant is 0.002845 1/M^4.

now we are given that trial is conducted at 100 C. so the equilibrium constant will be calculated as,

for trial E,

K [Coci) [C1-14 * Co(H20)2+1

we get

$K = \frac{[0.358091 M]}{[0.567634M]^4 * [0.641909M]} = 5.3733 M^{-4}$ ............. [ans.]

so, the equilibrium constant at 100 C is given as 5.3733 //M^4.