For the reaction - N 2 -3 H 2 - 2 NH 3 - - H-ve then-

For a general reaction aA(g) + bB(g) $⇋$ cC(g)+ dD(g), what is the relationship between the equilibrium constants $K_{c}$ and $K_{p}$ ? [assume $Δ$ n = $μ$ = (c + d) - (a+b) ]\)

Q. For the reaction,

P C I_{5} (g) ⇌ P C I_{3} (g) + C I_{2} (g)

K_{p}

and

K_{c}

are related as

K_{p} = K_{c} (R T)

K_{p}

for the reaction

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

400^{\circ} C

1.64 \times 10^{- 4}

Calculate

K_{C}

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For the reaction : N2+3H2⇌2NH3;ΔH=−ve then?

The correct option is C Kp=Kc(RT)−2N2(g)+3H2(g) ⇌2NH3(g);ΔH=−veΔng=2−(3+1)=−2Now, Kp=Kc(RT)Δng Kp=Kc(RT)−2

For the reaction : $N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}; Δ H = - v e$ then?

The correct option is C $K_{p} = K_{c} {(R T)}^{- 2}$
$N_{2} (g) + 3 H_{2} (g)$ $⇌ 2 N H_{3} (g); Δ H = - v e$

$Δ n_{g} = 2 - (3 + 1) = - 2$

Now, $K_{p} = K_{c} (R T)^{Δ n_{g}}$

$K_{p} = K_{c} (R T)^{- 2}$