Question

For the reaction SbCl5(g)? SbCl3(g) + Cl2(g),? ?
(SbCl5) = -334.34 kJ/mol?
(SbCl3) = -301.25 kJ/mol?
(SbCl5) = -394.34 kJ/mol?
(SbCl3) = -313.80 kJ/mol?

Will this reaction proceed spontaneously at 298 K and 1 atm pressure?


Can you please explain me how you got the answer? It is not making any sense to me

Answer 1

SbCl3(g) + Cl2(g)====>SbCl5(g)

I:0................0.62.............. 0.18
C:x...............x................... x
E:x...............0.62+x......... 0.18-x

Keq=[0.18-]/[x][0.62+x]=40

Solve using the quadratic equation and the answer should be about 0.0075.

So x=0.0075

Substitute x into the Equilibrium values in your ICEbox and you should get.

SbCl3(g)=0.0075 M
Cl2(g)=0.6275 M
SbCl5(g)=0.1725 M

Answer 2

(Delta G = Delta H - TDelta S)

Is (Delta G <0) reaction is spontaneous!

Answer 3

yes

Answer 4

SbCl3(g) + Cl2(g)====>SbCl5(g)

I:0................0.62.............. 0.18
C:x...............x................... x
E:x...............0.62+x......... 0.18-x

Keq=[0.18-]/[x][0.62+x]=40

Solve using the quadratic equation and the answer should be about 0.0075.

So x=0.0075

Substitute x into the Equilibrium values in your ICEbox and you should get.

SbCl3(g)=0.0075 M
Cl2(g)=0.6275 M
SbCl5(g)=0.1725 M