Question

relation between KP and KC for the reaction H2 + cl2 - 2HCL
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Answer 1

$<marquee>Bonjour !!$

⬛ $\boxed{\boxed{\huge{K_{p} = K_{c} RT^{Δn}}}}$

∆n = moles of reactant - moles of product

∆n = 0

$K_p = K_c RT^0\\ \\ K_p = K_c$

Answer 2

Both are equal $(K_p =K_c)$

Explanation:

• The reaction can be represented as: $H_2(g)+Cl_2(g)\rightarrow2HCl(g)$

The relation between the equilibrium constant $K_c$and$K_p$ for any reaction is $K_p =K_c(RT)^{\Delta_{ng}}$

where

• $\Delta_{ng}$is the difference in the number of moles of gaseous product and reactant

For, the given reaction,

• The number of moles of gaseous products $= 2$
• The number of moles of gaseous reactants $= 1+1 = 2$
• Hence, $\Delta_{ng} = 2 -(1+1) = 0$

Now, $K_p =K_c(RT)^{0}$

$K_p =K_c$