The correct relation between $K_{p}$ and $K_{c}$ for the given reaction is:
$H_{2} (g) + l_{1} (g) ⇌ 2 H l (g)$

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

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K_{p} = K_{c}

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\frac{K_{p}}{R T} = K_{c}

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

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Solution

The correct option is D $K_{p} = K_{c}$
Generally K $_{p} = K_{c} (R T)^{Δ n}$
$Δ n =$ no. of mole of products - no. of moles of reactants.
Here no. of product moles $= 2$
no. of reactant moles $= 2$
So $Δ n = 2 - 2 = 0$
$K_{p} = K_{c} (R T)^{0} = K_{c}$
So for this reaction, $K_{p} = K_{c}$

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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) ]\)

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If true enter 1, else enter 0.

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The correct relation between Kp and Kc for the given reaction is:H2(g)+l1(g)⇌2Hl(g)

The correct option is D Kp=KcGenerally Kp=Kc(RT)ΔnΔn= no. of mole of products - no. of moles of reactants. Here no. of product moles =2no. of reactant moles =2So Δn=2−2=0Kp=Kc(RT)0=KcSo for this reaction, Kp=Kc

The correct relation between $K_{p}$ and $K_{c}$ for the given reaction is:
$H_{2} (g) + l_{1} (g) ⇌ 2 H l (g)$

The correct option is D $K_{p} = K_{c}$
Generally K $_{p} = K_{c} (R T)^{Δ n}$
$Δ n =$ no. of mole of products - no. of moles of reactants.
Here no. of product moles $= 2$
no. of reactant moles $= 2$
So $Δ n = 2 - 2 = 0$
$K_{p} = K_{c} (R T)^{0} = K_{c}$
So for this reaction, $K_{p} = K_{c}$