The equilibrium constant, Kp for the reaction 2SO2(g) + O2(g) leftharpoons 2SO3(g) is 4.0 atm-1 at 1000 K. What would be the partial pressure of O2 if at equilibrium the amount of SO2 and SO3 is the same?

Question

The equilibrium constant, Kp for the reaction 2SO2(g) + O2(g) leftharpoons 2SO3(g) is 4.0 atm-1 at 1000 K. What would be the partial pressure of O2 if at equilibrium the amount of SO2 and SO3 is the same? (A) 16.0 atm (B) 0.25 atm (C) 1 atm (D) 0.75 atm

Tardigrade Admin · Accepted Answer

2SO2 (g) + O2 leftharpoons 2SO3(g) Kp = 4.0 atm-2 Kp =((SO3)2/(SO2)2 (O2)) Given that at equilibrium the amount of SO2 and SO3 is the same so ((SO3)2/(SO2)2 (O2)) =4 ⇒ [O2] = (1/2) =0.25 atm

Q. The equilibrium constant, Kp​ for the reaction 2SO2​(g)+O2​(g)⇌2SO3​(g) is 4.0atm−1 at 1000K. What would be the partial pressure of O2​ if at equilibrium the amount of SO2​ and SO3​ is the same?

16.0 atm

0.25 atm

1 atm

0.75 atm

2SO2​(g)+O2​⇌2SO3​(g) Kp​=4.0atm−2 Kp​=(SO2​)2(O2​)(SO3​)2​ Given that at equilibrium the amount of SO2​ and SO3​ is the same so (SO2​)2(O2​)(SO3​)2​=4⇒[O2​]=21​=0.25 atm

Q. The equilibrium constant, $K_{p}$ for the reaction
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
is $4.0 a t m^{- 1}$ at $1000 K .$ What would be the partial pressure of $O_{2}$ if at equilibrium the amount of $S O_{2}$ and $S O_{3}$ is the same?