Question

The equilibrium between NO2 and N2O4 can be described by the following equation:

2NO2(g) ⇌ N2O4(g) Kp = 7.0

If a sealed flask contains 1.5 atm of NO2 and 14.2 atm of N2O4. Calculate the value of Kc for the reaction,

Answer 1

The question is incomplete, here is the complete question:

The equilibrium between NO₂ and N₂O₄ can be described by the following equation:

2NO₂(g) ⇌ N₂O₄(g); Kp = 7.0

If a sealed flask contains 1.5 atm of NO₂ and 14.2 atm of N₂O₄. Calculate the value of Kc for the reaction at 25°C

Answer: The value of $K_c$ for given reaction is 171.3

Explanation:

For the given chemical equation:

$2NO_2(g)\rightleftharpoons N_2O_4(g)$

Relation of $K_p\text{ with }K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta n_{g}$

Where,

$K_p$ = equilibrium constant in terms of partial pressure = 7.0

$K_c$ = equilibrium constant in terms of concentration

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

T = temperature = $25^oC=25+273=298K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=1-2=-1$

Putting values in above equation, we get:

$7.0=K_c\times (0.0821\times 298)^{-1}\\\\K_c=171.3$

Hence, the value of $K_c$ for given reaction is 171.3