2) For the equilibrium: 2 SO3(g) <=> 02(g) + 2 SO2(g) Kp = 0.269 at 625...

Question

Question

2) For the equilibrium: 2 SO3(g) <=> 02(g) + 2 SO2(g) Kp = 0.269 at 625 °C What is Ke at this temperature? Kp = K[R R = 0.082

science chemistry

Answer 1

Answer #1

sol:-

data provided in the question is :-

Reaction is :-

2503(g) = 02(g) +2502(8)

$\Delta{n}$ = (sum of mole of product) - (sum of mole of reactant)

Δη = (2+1) – (2) = 1

put all value in formula :- then,

0.269 = K (0.08206 x 898) Kc = (0.269)/(0.08206 x 898)

= (0.269)/(73.68988)

= 0.00365

Ke=0.00365

Answer 2

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