Answer to Question #128284 in Inorganic Chemistry for rishabh

Question

Consider the following reversible reaction occurring in a 1.00 L enclosed vessel:
(NH4)2SO4(s) 2NH3(g) + H2SO4(g)
Given that the value of the equilibrium constant, Kc, of this reaction is 2.57 × 10-5, calculate the molar concentrations of NH3(g) and H2SO4(g) at equilibrium. No NH3(g) and H2SO4(g) were present initially.

Answer 1

In the equilibrium constant we take into account only gaseous components. Thus, equilibrium constant can be writen as:

$K_{c} = c^{2}(NH_{3})\times c(H_{2}SO_{4})$

$c(NH_{3}) = 2c(H_{2}SO_{4})$

$(2c)^{2}c=K_{c}=2.57\times10^{-5}$

$c(H_{2}SO_{4})= (\frac{2.57\times10^{-5}}{4})^{\frac{1}{3}}=0.0186\frac{mol}{L}$

$c(NH_{3})=0.0186\times 2 = 0.0372\frac{mol}{L}$