At 450K- displaystyle - K - p -2-0times - 10 - 10 -bar the given reaction equilibrium-displaystyle 2- SO - 2 -left- g right- - O - 2 -left- g right- rightleftharpoons 2- SO - 3 -left- g right- What is displaystyle - K - c - this temperature-

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Toppr Admin · Accepted Answer

-x394-n-nproducts-x2212-nreactants-2-x2212-2-1-x2212-1we know that-xA0-Kp-Kc-RT-x394-n-xA0-x2234-Kc-Kp-RT-x2212-x394-n-2-0-xD7-1010-xD7-0-0831-xD7-450-1-7-48-xD7-1011M-mol

At 450K, $K_{p} = 2.0 \times 10^{10}$ /bar for the given reaction at equilibrium.

$2 {S O}_{2} (g) + O_{2} (g) ⇌ 2 {S O}_{3} (g)$

What is $K_{c}$ at this temperature?

$Δ n = n_{p r o d u c t s} - n_{r e a c t a n t s} = 2 - (2 + 1) = - 1$

we know that $K_{p} = K_{c} (R T)^{Δ n}$

$∴ K_{c} = K_{p} (R T)^{- Δ n} = (2.0 \times 10^{10}) \times (0.0831 \times 450)^{1} = 7.48 \times 10^{11} M / m o l$

At 450K, Kp=2.0×1010/bar for the given reaction at equilibrium.2SO2(g)+O2(g)⇌2SO3(g)What is Kc at this temperature?

Δn=nproducts−nreactants=2−(2+1)=−1we know that Kp=Kc(RT)Δn ∴Kc=Kp(RT)−Δn=(2.0×1010)×(0.0831×450)1=7.48×1011M/mol

At 450K, $K_{p} = 2.0 \times 10^{10}$ /bar for the given reaction at equilibrium.

$2 {S O}_{2} (g) + O_{2} (g) ⇌ 2 {S O}_{3} (g)$

What is $K_{c}$ at this temperature?

$Δ n = n_{p r o d u c t s} - n_{r e a c t a n t s} = 2 - (2 + 1) = - 1$

we know that $K_{p} = K_{c} (R T)^{Δ n}$

$∴ K_{c} = K_{p} (R T)^{- Δ n} = (2.0 \times 10^{10}) \times (0.0831 \times 450)^{1} = 7.48 \times 10^{11} M / m o l$