Find out the value of K_C for the following r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The relationship between $K_P$ and $K_C$ is given by the formula: $K_P = K_C(RT)^{\Delta n_g}$.For the reaction $2NOCl_{(g)} \rightleftharpoons 2N

Vedclass · Accepted Answer

$2 	imes 10^{11} \ mol \ L^{-1}$

Vedclass · Answer

(C) The relationship between $K_P$ and $K_C$ is given by the formula: $K_P = K_C(RT)^{\Delta n_g}$.For the reaction $2NOCl_{(g)} ightleftharpoons 2NO_{(g)} + Cl_{2(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta n_g = (2 + 1) - 2 = 1$.Given values are $K_P = 8 	imes 10^{12} \ atm$,$T = 500 \ K$,and $R = 0.08 \ L \ atm \ mol^{-1} \ K^{-1}$.Substituting these values into the formula: $8 	imes 10^{12} = K_C 	imes (0.08 	imes 500)^1$.$8 	imes 10^{12} = K_C 	imes 40$.$K_C = \frac{8 	imes 10^{12}}{40} = 0.2 	imes 10^{12} = 2 	imes 10^{11} \ mol \ L^{-1}$.

Find out the value of $K_C$ for the following reaction from the value of $K_P$
$2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$
[Given : $K_P = 8 \times 10^{12} \ atm$ at $500 \ K$ use $R = 0.08 \ L \ atm \ mol^{-1} \ K^{-1}$ ]

Similar Questions

Which one of the following is heterosporous?

The figure shows the volume $V$ versus temperature $T$ graphs for a certain mass of a perfect gas at two constant pressures of $P_1$ and $P_2$. What inference can you draw from the graphs?

In Young's double slit experiment,one of the slits is wider than the other,so that the amplitude of the light from one slit is double that from the other slit. If $I_m$ be the maximum intensity,the resultant intensity when they interfere at phase difference $\phi$ is given by:

The number of linkage groups in $Pisum$ $sativum$ is:

Hydrophilic sols are

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find out the value of $K_C$ for the following reaction from the value of $K_P$$2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$[Given : $K_P = 8 \times 10^{12} \ atm$ at $500 \ K$ use $R = 0.08 \ L \ atm \ mol^{-1} \ K^{-1}$ ]