If you find that (CoC142 (aq)] = 0.306 M, [Co(H20).2+(aq)] = 0.76 M, and (Cl-(aq)] =...

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Question

Answer 1

Answer #1

consider the following equilibrium

Co(H₂O)₆²⁺(aq) + 4 Cl^-(aq) $\rightleftharpoons$ CoCl₄^2-(aq) + 6 H₂O(g)

K_equilibrium =[CoCl₄^2-][H₂O]⁶/[Co(H₂O)₆²⁺][Cl^-]⁴

[H₂O] is taken as 1

K_equilibrium =[CoCl₄^2-]/[Co(H₂O)₆²⁺][Cl^-]⁴

K_equilibrium =0.306/0.76*(0.193)⁴

K_equilibrium =0.306/(1.05 *10^-3)

K_equilibrium =290.187

temperature T =4⁰C =273 +4 =277K

we know

$AG = -RTln Kequilibrium$

$AG = -8.314J K-1 * 277K * In 290.187$

$AG = -8.314J K-1 * 277K *5.67$

$AG = -8.314J K-1 * 277K *5.67 = -13.059 * 1031$

$AG = -13.059kJ$

Answer 2

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