K P is how ma...

Question

$K_{P}$ is how many times equal to $K_{c}$ for the given reaction?
$N_{2} (g) + 3 H_{2} (g) ⇌ {2 N H}_{3} (g)$

\frac{1}{R^{2} T^{2}}

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R^{2} T^{2}

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\frac{R}{T}

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R T

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Solution

The correct option is C $\frac{1}{R^{2} T^{2}}$
As we know that,
$K_{p} = K_{c} {(R T)}^{Δ n_{g}}$
where as,
$Δ n_{g} =$ Sum of stoichiometric coefficients of product - Sum of stoichiometric coefficients of reactant $= n_{p} - n_{r}$
Therefore, for the reaction,
${N_{2}}_{(g)} + 3 {H_{2}}_{(g)} ⇌ 2 {N H_{3}}_{(g)}$
$Δ n_{g} = 2 - (1 + 3) = 2 - 4 = - 2$
$∴ K_{p} = K_{c} {(R T)}^{- 2}$
$\Rightarrow K_{p} = \frac{1}{R^{2} T^{2}} K_{c}$
Hence, for the given reaction, $K_{p}$ is $\frac{1}{R^{2} T^{2}}$ times equal to $K_{c}$ .

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