Question

For which of the following reaction is kp = kc
(A) H2(g) + 12(g) - 2HI(g)
(B)2N204(9) - 4NO2(g)
(C) N2(g) + 3H2(g) 2NH3(g)
(D) H2(g) + Cl2(g) + 2HCl(g)​

Answer 1

Answer:

The reactions in option A and option D has  $K_p\ =\ K_c$ ·

$(A)\ \ H_2_{(g)}\ +\ I_2_{(g)} \ --->\ 2\ HI_{(g)}$

$(D)\ \ H_2_{(g)} \ +\ Cl_2_{(g)} \ --->\ 2\ HCl_{(g)}$

Explanation:

We know that both $K_p$ and $K_c$ are the equilibrium constants, and the relation that connects both of them follows,

   $K_p\ =\ K_c\ (R\ T)^{\Delta n}$

where,

$K_p$ and $K_c$ are the equilibrium constants·

R  $=$ The universal gas constant

$\Delta n\ =$ The No of moles of the product $-$ The No of moles of the reactant

When the change in No of moles $=\ 0$ the reaction will have a relationship that  $K_p\ =\ K_c$·

So, to find out which of the following reaction has  $K_p\ =\ K_c$  we should check which has $\Delta n\ =$ $0.$

$(A)\ \ H_2_{(g)}\ +\ I_2_{(g)} \ --->\ 2\ HI_{(g)}$

Here,

The No of moles of the product $=\ 2$

The No of moles of the reactant $=$ $1\ +\ 1\ =\ 2$

∴  $\Delta n\ =\ 2\ -\ 2\ =\ 0$

$(B)\ \ 2\ N_2O_4_{(g)}\ --->\ 4\ NO_2_{(g)}$

Here,

The No of moles of the product $=$ $4$

The No of moles of the reactant $=$ $2$

∴  $\Delta n\ =\ 4\ -\ 2\ =\ 2$

$(C)\ \ N_2_{(g)} \ +\ 3\ H_2_{(g)} \ --->\ 2\ NH_3_{(g)}$

Here,

The No of moles of the product $=$ $2$

The No of moles of the reactant $=$ $3\ +\ 1\ =$ $4$

∴  $\Delta n\ =\ 2\ -\ 4\ =\ -2$

$(D)\ \ H_2_{(g)} \ +\ Cl_2_{(g)} \ --->\ 2\ HCl_{(g)}$

Here,

The No of moles of the product $=$ $2$

The No of moles of the reactant $=$ $1\ +\ 1\ =$  $2$

∴ $\Delta n\ =\ 2\ -\ 2\ =\ 0$

From the above information option, A and option D has $\Delta n\ =$ $0.$

Therefore,

The reactions in option A and option D has  $K_p\ =\ K_c$·