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# Answer to Question #128284 in Inorganic Chemistry for rishabh

Question #128284
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.
1
2020-08-05T05:09:24-0400

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}"

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