Consider the following reaction where Kp-625 at 298 K: 2NO(g) + Br2(g)2NOBr(g) I

Question

Consider the following reaction where Kp-625 at 298 K: 2NO(g) + Br2(g)2NOBr(g) If the three gases are mixed in a rigid container at 298 K so that the partial pressure of each gas is initially one atm, what will happen? -B 1. A reaction will occur in which NOBr(g) is produced. -D 2. Kp will increase. 3. A reaction will occur in which NO is produced. 4. Q is greater than K. 5. The reaction is at equilibrium. No further reaction will occur. - Submit Answer Retry Entire Group No more group attempts remain

Explanation / Answer

   2NO(g) + Br2(g) <-----> 2NOBr(g)

Reaction quotient(Qp) = 1^2/(1^2*1) = 1

   Kp = 6.25

Qp < Kp , backward reaction takes place.

1. False

2. False

3. True

4. False

5. False.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---