Question

2 questions, need help!

When the oxide of generic metal M is heated at 25.0 °C, a negligible amount of M is produced kJ AG" 289.0- mol MO2(s) M(s)+ 02(g) When this reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous What is the chemical equation of this coupled process? Show that the reaction is in equilibrium. Include physical states and represent graphite as C(s) chemical equation: MO,(s)C(s)=M(s) + CO2(g) What is the thermodynamic equilbrium constant for the coupled reaction? 2.98461x1018 K =

Answer 1

Answer: - Given thats Mod(s) = M(S) + 02 (9) mol AG' =289.0 kati And M is heated at 25°c The chemical equation is Moa (5) + C(s) = M(S) +009 (9) Now, Take > Mg(s) M(S) + Og (9) AG=289.0 kg/mol

And fore(s) + 09 (9) Com (G) 442 > - 324.4 leg Imol Add the both then we will Mog(sht C(s) + O2 (9) M(S) HO2 (9) Now, AG will be AG = AGIH AGO Now, Substitute those values then. Ano = 989 + (- 394.4) 460 = - 105.4 kg/mol

so AG z - 105400 g/mol As given temperature (T) = 25 in kelvin - T=298.15k And R= 8.314 J/mol k. Res molar gas constant Now, we know that k= Ep (-A Go /RT)

Now Substitute those values. K= Sup CC-105400) - (-105400) 8.314 x 298.15 K = Exp (3.1314x298115) r 105400 105400 ka Exp (2478.8191 ) k = Exp (42.520) k = 2.94% 1018 The answer is [ k= 2.92X 108 /

A (aq) + B (aq) c (aq) aA + bB ²oC k=cosc CAJCBJb T=95°c > 298.15% Standard free Energy change = -4:53 kolme Now, 46° = - RT In keq 8.314 -4.53k= -8.314 (298) In keq -4.53k ln kea 68.314) (298)

In keq = 1.828 beq: e 1:828 keq = 6.22 c (aq) A caq) + Blaq) = 0.3 0. 4 60.3 -) Cos4-x) o Enitial Quilibacium e [c [A] [B] 6.22= (0.30-2) (0.40x) 6.22€0.30-x) (0.4-2)) = x

6.22 (0.12 -0.982 -0.4x + xy) = x 2-0.74 +x2) = 2 0.74-4.35 +6.2222 => 6.2222-4.35X +0.74 = x - bt 5 b² nac - Qa -6-4.35) + 36-4.35)? – 466-22) (0.44) 2(6.22) 4.351 0.71 12.44 0.4 & 0.29

QÅ x=0.gq Equillibarium Concentration of (472 0.30 - 0.29 - = 0.01 M "! [B] = 0,4-0,29 20.11 M 11 " [c] = x = 0.29 M À & B' but moreč There would be less 1 & f x=0.40 then. [A] = 0.30-0.4 (BJ = 0.4-0.4 = 0 [c] = x=0.40