Question

2.1 kJ/mol. The following reaction has a value of AGⓇ = CH3Br + H2S CH,SH HBr a) Calculate key at room temperature (25°C) for this reaction as written. b) Starting with 1 M solution each of CH3Br and H2S, calculate the final concentration of all four species at equilibrium. Please be thorough in your presentation AGⓇ = -2.303RT(log kea); at 25°C, RT= 2478 J/mol or 2.48 kJ/mol
the value on top is -2.1kj/mol

Answer 1

a Go= -2.1 kJ molt 160 = -2°303 RT (lug key) Reg- equillebrium constant RT = 2.48 KJ molt - -2.1 KJ Imol: -2.303 X 2.48 kJ molt x - log keq keq - 2033 Hence, keg at room temperature (25°C) is 2133 b) [cHg Bo ] = equillebnum concentration [H2S] = equillibrum concentration [CH3 SH] = egerillihn um concentration THB) = equillibrum concentration of of Chiz Br of ths of CH₂SH Hor keq = [CH3SH] [MBOL = 233 0 [(Fts By ] [H₂5] ICE Table concentration (M) CHABE + HS - CH3SH & HBO c x ox 1-3 - 1-3 x x E

= 2.33 =) keq = xxx (1-2)(1-) + 1x 12 = 2.33 Taking square root on both side we get X = 11526 x= 0.604 M [CH3 Bo] = (1 - x0) M = (1-0.604) M = 0.396M [H2S] = (1-0) M = (1-0.604) M = 0.396 M [CH3 SH] = 1 = 0.604 M ure [HBr J = x = 0.604 M. Pleue Rate