For reaction: SO 2( g )+(1/2) O 2( g )= SO 3( g ) K p =2 × 1012 at 27° C and 1 atm pressure. The K c for the same reaction is × 1013. (Nearest integer) (Given R =0.082 L atm K -1 mol -1 )

Question

For reaction: SO 2( g )+(1/2) O 2( g )= SO 3( g ) K p =2 × 1012 at 27° C and 1 atm pressure. The K c for the same reaction is × 1013. (Nearest integer) (Given R =0.082 L atm K -1 mol -1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

SO 2( g )+(1/2) O 2( g ) leftharpoons SO 3( g ) K p =2 × 1012 text at 300 K K p = K C ×( RT )Δ n g 2 × 1012= K c ×(0.082 × 300)-1 / 2 K C =9.92 × 1012 K C =0.992 × 1013 = 1

Q. For reaction : $S O_{2} (g) + \frac{1}{2} O_{2} (g) = S O_{3} (g)$ $K_{p} = 2 \times 1 0^{12}$ at $2 7^{\circ} C$ and 1 atm pressure. The $K_{c}$ for the same reaction is ____ $\times 1 0^{13}$ . (Nearest integer) (Given $R = 0.082 L a t m K^{- 1} m o l^{- 1}$ )

$S O_{2 (g)} + \frac{1}{2} O_{2 (g)} ⇌ S O_{3 (g)}$
$K_{p} = 2 \times 1 0^{12} at 300 K$
$K_{p} = K_{C} \times (RT)^{Δ n_{g}}$
$2 \times 1 0^{12} = K_{c} \times (0.082 \times 300)^{- 1/2}$
$K_{C} = 9.92 \times 1 0^{12}$
$K_{C} = 0.992 \times 1 0^{13}$
$= 1$

Q. For reaction : SO2​(g)+21​O2​(g)=SO3​(g) Kp​=2×1012 at 27∘C and 1 atm pressure. The Kc​ for the same reaction is ____×1013. (Nearest integer) (Given R=0.082LatmK−1mol−1 )

SO2(g)​+21​O2(g)​⇌SO3(g)​ Kp​=2×1012 at 300K Kp​=KC​×(RT)Δng​ 2×1012=Kc​×(0.082×300)−1/2 KC​=9.92×1012 KC​=0.992×1013 =1

Q. For reaction : $S O_{2} (g) + \frac{1}{2} O_{2} (g) = S O_{3} (g)$ $K_{p} = 2 \times 1 0^{12}$ at $2 7^{\circ} C$ and 1 atm pressure. The $K_{c}$ for the same reaction is ____ $\times 1 0^{13}$ . (Nearest integer) (Given $R = 0.082 L a t m K^{- 1} m o l^{- 1}$ )

$S O_{2 (g)} + \frac{1}{2} O_{2 (g)} ⇌ S O_{3 (g)}$
$K_{p} = 2 \times 1 0^{12} at 300 K$
$K_{p} = K_{C} \times (RT)^{Δ n_{g}}$
$2 \times 1 0^{12} = K_{c} \times (0.082 \times 300)^{- 1/2}$
$K_{C} = 9.92 \times 1 0^{12}$
$K_{C} = 0.992 \times 1 0^{13}$
$= 1$