At T (K ), the K_p for the reaction A_2B_6 (g ) A_2B_4 (g )+ B_2 (g ) is 0.04 atm. The equilibrium pressure (in atm) of (A_2B_6 )g ) when it is placed in a flask at 4 atm pressure and allowed to come…

Question

At T (K ), the K_p for the reaction A_2B_6 (g ) A_2B_4 (g )+ B_2 (g ) is 0.04 atm. The equilibrium pressure (in atm) of (A_2B_6 )g ) when it is placed in a flask at 4 atm pressure and allowed to come to above equilibrium is

Accepted Answer

Correct answer: C. (c) A_2 B_6 ( g ) ⇌A_2 B_4 ( g )+B_2 ( g ) Initial p=4 atm 0.0 0.0 Equilibrium 4-4 p =4 p =4 =4 (1- )K_p= P (A_2 B_4 ) P (B_2 )P (A_2 B_6 )=0.04 atm = (4 ) (4 )4 (1- )= 4 ^21- ≈4 ^2 ⇒ =0.1 Therefore, P (A_2 B_6 ) at equilibrium =4 (1- ) =4 (1-0.1 )=3.62

$At T (K), the K_{p} for the reaction A_{2} B_{6} (g) \to A_{2} B_{4} (g) + B_{2} (g) is 0.04 atm . The equilibrium pressure (in atm) of (A_{2} B_{6}) g) when it is placed in a flask at 4 atm pressure and allowed to come to above equilibrium is$

Solution

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