Question

Some SbClş is allowed to dissociate into SbCh, and Chat 521 K. At equilibrium, [SHCl.) -0.279 M, and [SbC1] - [C13] - 8.34 102 M. Additional ShCh is added so that SC -0.147 M and the system is allowed to once again reach equilibrium SbCis(g) – SbC13(8) - ChE) K-2.5010 at 521 K (a) In which direction will the reaction proceed to reach equilibrium? (6) What are the new concentrations of reactants and products after the system reaches equilibrium? [SbCls) - [SC] - [Cl] -

Answer 1

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PART (a):

Given reaction:

+ SbCl; (g) At eqbm: 0.279 M After SbC1z = SbCl; (g) 0.0834 M Cl2 (g) 0.0834 M ;K=0.025 added : 0.279 M 0.147 M 0.0834 M ;Q=- (0.147) X (0.0834) ==0.044 (0.279) .:Q (reaction quotient) > K (equilibrium constant). Hence, Backward reacrtion will occur. That towards the left.

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PART (b):

+ From ICE Table: SbCl, (g) Initial: 0.279 M Change: +x At eqbm : 0.279+x Cl, (g) 0.0834 M ;K=0.025 = SbC1z (g) 0.147 M -X 0.147 - X -X 0.0834 - x

$\therefore K=\frac{(0.147-x).(0.0834-x)}{(0.279+x)}=0.025$

$\therefore (0.147-x).(0.0834-x)=(0.279+x)\times0.025$

$\therefore x^{2}-0.2554x+0.0052848=0$

Solving for x, we get,

$x=0.0227\;M$

$\therefore [SbCl_{5}]=0.279+x=0.279\;M+0.0227\;M\mathbf{=0.3017\;M}$

$\therefore [SbCl_{3}]=0.147-x=0.147\;M-0.0227\;M\mathbf{=0.1243\;M}$

$\therefore [Cl_{2}]=0.0834-x=0.0834\;M-0.0227\;M\mathbf{=0.0607\;M}$

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