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Relation between Kp and Kc

For the equil...

Question

For the equilibrium reaction $3 F e (s) + 4 H_{2} O (g) ⇋ F e_{3} O_{4} (s) + 4 H_{2} (g)$ , the relationship between $K_{p}$ and $K_{c}$ is:

K_{p} > K_{c}

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K_{p} < K_{c}

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

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K_{p} = K_{c}

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Solution

The correct option is C $K_{p} = K_{c}$
The relationship between $K_{p}$ and $K_{c}$ is as follows:
$K_{p} = K_{c} (R T)^{Δ n}$
When $Δ n = 0$ , the term $(R T)^{Δ n}$ becomes $(R T)^{0} = 1$
Hence, $K_{p} = K_{c}$
This is true for the reaction $3 F e (s) + 4 H_{2} O (g) \Leftrightarrow F e_{3} O_{4} (s) + 4 H_{2} (g)$ for which $Δ n = 4 - 4 = 0$
Here, only the number of moles of gaseous substances are considered.

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For the equilibrium reaction 3Fe(s)+4H2O(g)⇋Fe3O4(s)+4H2(g), the relationship between Kp and Kc is:

The correct option is C Kp=KcThe relationship between Kp and Kc is as follows:Kp=Kc(RT)ΔnWhen Δn=0, the term (RT)Δn becomes (RT)0=1Hence, Kp=KcThis is true for the reaction 3Fe(s)+4H2O(g)⇔Fe3O4(s)+4H2(g) for which Δn=4−4=0 Here, only the number of moles of gaseous substances are considered.

For the equilibrium reaction $3 F e (s) + 4 H_{2} O (g) ⇋ F e_{3} O_{4} (s) + 4 H_{2} (g)$ , the relationship between $K_{p}$ and $K_{c}$ is: