In the synthesis of Ammonia by Haber-s Processs- N 2 g -3 H 2 g - 2 NH 3 g at 500- C- the value of K p was found to be 0-5 - Calculate the value of K c-

K_p = K_c (RT)^Δ n where, K_p = 0.5 (given) R = 0.083 bar dn^3 T=500^∘ C = 500+273 = 773 k Δ n = 2 4 = 2 k_c = k_p/(RT)^Δ n = 0.5/(0.083 × 773)^ 2 =0.5 (0.083 ...

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Standard XII

Chemistry

Hess' Law

In the synthe...

Question

In the synthesis of Ammonia by Haber's Processs.
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ at $500^{\circ}$ C, the value of $K_{p}$ was found to be 0.5. Calculate the value of $K_{c}$ .

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Solution

$K_{p} = K_{c} (R T)^{Δ n}$
where,
$K_{p} = 0.5$ (given)
$R = 0.083 b a r d n^{3}$
$T = 500^{\circ} C = 500 + 273 = 773 k$
$Δ n = 2 - 4 = - 2$
$k_{c} = \frac{k_{p}}{(R T)^{Δ n}} = \frac{0.5}{(0.083 \times 773)^{- 2}}$
$= 0.5 (0.083 \times 773)^{2}$
$= 2063.15$

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In the synthesis of Ammonia by Haber's Processs. N2(g)+3H2(g)⇌2NH3(g) at 500∘C, the value of Kp was found to be 0.5. Calculate the value of Kc.

Kp=Kc(RT)Δn where, Kp=0.5 (given) R=0.083 bar dn3 T=500∘ C=500+273=773 k Δn=2−4=−2 kc=kp(RT)Δn=0.5(0.083×773)−2 =0.5(0.083×773)2 =2063.15

In the synthesis of Ammonia by Haber's Processs.
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ at $500^{\circ}$ C, the value of $K_{p}$ was found to be 0.5. Calculate the value of $K_{c}$ .

$K_{p} = K_{c} (R T)^{Δ n}$
where,
$K_{p} = 0.5$ (given)
$R = 0.083 b a r d n^{3}$
$T = 500^{\circ} C = 500 + 273 = 773 k$
$Δ n = 2 - 4 = - 2$
$k_{c} = \frac{k_{p}}{(R T)^{Δ n}} = \frac{0.5}{(0.083 \times 773)^{- 2}}$
$= 0.5 (0.083 \times 773)^{2}$
$= 2063.15$