Question

Consider the following reaction: 2NOBr(g) 2NO(g) + Br2(g) If 0.193 moles of NOBr, 0.210 moles of NO, and 0.293 moles of Br2 are at equilibrium in a 12.0 L container at 413 K, the value of the equilibrium constant, Kp, is .

Answer 1

Equilibrium constant Kp=

Volume of container=V=12.0 L

Temperature T=413 K

Gas constant R=0.0821 L atm/molK

Number of moles of NO at equilibrium=n_{​​​​​​NO}​=0.210 mol

Number of moles of Br₂ at equilibrium=n_Br2=0.293 mol

Number of moles of NOBr at equilibrium=n_{​​​​​​NOBr}=0.193 mol

p_NO=Partial pressure of NO=n_NORT/V

p_Br2=Partial pressure of Br₂=nBr2RT/V

p_NOBr=Partial pressure of NOBr=n_NOBrRT/V

===0.98 atm

So the equilibrium constant Kp=0.98 atm