KpKc for the following reaction will be COg - 12 O2g - CO2g

Δ n_g = n_p n_r = 1 32= 12 n_p is number of moles of gaseous compound in product side nbsp; n_r is number of moles of gaseous compound in reactant side nb ...

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Byju's Answer

Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

KpKc for the ...

Question

$\frac{K_{p}}{K_{c}}$ for the following reaction will be
$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$

R T

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\frac{1}{R T}

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\frac{1}{\sqrt{R T}}

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\frac{R T}{2}

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Solution

The correct option is C $\frac{1}{\sqrt{R T}}$
$Δ n_{g} = n_{p} - n_{r} = 1 - \frac{3}{2} = - \frac{1}{2}$
$n_{p}$ is number of moles of gaseous compound in product side
$n_{r}$ is number of moles of gaseous compound in reactant side
$Δ n_{g} = - \frac{1}{2}$
Hence, $K_{p} = K_{c} (R T)^{- \frac{1}{2}}$
$\frac{K_{p}}{K_{c}} = \frac{1}{(R T)^{1 / 2}} = \frac{1}{\sqrt{R T}}$
Hence, option(c) is correct.

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KpKc for the following reaction will be CO(g)+12O2(g)→CO2(g)

The correct option is C 1√RTΔng=np−nr=1−32=−12 np is number of moles of gaseous compound in product side nr is number of moles of gaseous compound in reactant side Δng=−12 Hence, Kp=Kc(RT)−12 KpKc=1(RT)1/2=1√RT Hence, option(c) is correct.

$\frac{K_{p}}{K_{c}}$ for the following reaction will be
$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$