At 400.°C, K :0.250 for Br2(g) + 12(g)% 21 Br(g). Assume each gas behaves ideall

Question

At 400.°C, K :0.250 for Br2(g) + 12(g)% 21 Br(g). Assume each gas behaves ideally at this temperature. a. If enough gaseous Br2, gaseous l2, and gaseous IBr are placed in an initially empty 100. L container at 400. C to give partial pressures of o.0500 atm, 0.0500 atm, and 0.0800 atm for Br2, 12, and IBr, respectively, what is the partial pressure of each substance at equilibrium at 400. C? If only gaseous IBr is placed in the initially empty 100. L container at 400°C, giving a pressure of 0.1800 atm in the container, what are the partial pressures of Br2, 12, and IBr at equilibrium at 400. C for the reaction shown above? Does the similarity between your answers to Parts b and c surprise you? Why or why not? b. c.

Explanation / Answer

For the reaction I2(g)+ Br2(g)<-->2IBr,

Q= reaction coefficient = [IBr]2/[I2][Br2]=0.08*0.08/(0.05*0.05)=2.56>K, so the reaction has to proceed backward towards the reactants side to reduce Q and so as to reach equilibrium.

Let PI2, PBr2 and PIBr are the partial pressures of I2, Br2 and IBr at equilibrium

Let P= Additional partial pressure of I2 to reach equilibrium =

PI2=0.05+P =PBr2, PIBr= 0.08-2P

K= (0.08-2P)2/ (0.05+P)2= 0.25, taking square root, the equation becomes

(0.08-2P)/(0.05+P)= 0.5, 0.08-2P=0.5*(0.05+P), 0.08-2P=0.5*0.05+0.5P

2.5P= 0.08-0.5*0.05=0.055 and P=0.022 atm

So at equilibrium, PI2= PBr2=0.05+0.022=0.072 atm and PIBr= 0.08-2*0.022 =0.036 atm

2. only IBr2 is placed at P=0.18 atm Q= 0.18*0.18/0 = infinity >K , so the reaction has to again proceeds back ward so as to reach equilibrium at which Q=K

At Equilibrium, PI2= PBr2= P, and PIBR=0.18-2P

K= (0.18-2P)2/ P2= 0.25, Taking square root, (0.18-2P)/P= 0.5

0.18-2P=0.5P, 2.5P=0.18, P= 0.18/2.5=0.072 atm, PIBR= 0.18-2*0.072=0.036 atm

In both the cases, the reaction coefficient is >K, so the reaction has to proceed towards the reactants side. In case -1, since the initial partial pressure of IBr is less compared to case-2. There is already some pressure of I2 and Br2 in case-1. Here also Q>K, leading to the reaction proceeding in the forward direction.

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