Consider the gas-phase decomposition of NOBr: 2 NOBr (g) 2 NO (g) + Br2 (g) 1. W
ID: 679203 • Letter: C
Question
Consider the gas-phase decomposition of NOBr: 2 NOBr (g) 2 NO (g) + Br2 (g)1. Write the equilibrium constant expressions Kc and Kp for thisreaction.
2. 0.0200 moles of NOBr are added to an empty 1.00-L flask andallowed to reach equilibrium at 300 K.
When equilibrium is reached, the total pressure in the flask is0.588 atm. Calculate the equilibrium
constant Kc for this reaction at 300 K.
3. Determine the value of Kp for this reaction at 300 K.
4. A 2.00-L flask contains 0.20 mol NOBr, 0.15 mol NO, and 0.30 molBr2 at 300 K. Calculate the
reaction quotient Q, and determine if NOBr will be produced orconsumed under these conditions.
Explanation / Answer
1. Kc=[NO]2[Br2]/[NOBr]2 Kp = P2NO x PBr2 /P2NOBrAll products and reactants aregases, therefore they all have partial pressures and changingconcentrations and are in both equilibrium expressions. 2. .0200 moles NOBr Using PV = nRT to find the partial pressure of NOBr at thebeginning: P = .02 mole x .0821 x 300K/ 1.00L = .4926 atm In order to reach equilibrium, NOBr will decompose and losemolecules, and therefore pressure at the same proportion. According to the balanced equation, NOBr will lose 2y atm in orderto reach equilibrium, NO will gain 2y atm, and Br2 will gain yatm. The equilibrium value for the atm of each will be: NOBr: ..4926 - 2y NO: 2y Br2: y The total pressure at equilibrum is .588 atm, .588 atm =.4926 - 2y +2y +y y = .0954 atm, so we have NOBr at .3018 atm, NO at .1908 atm, andBr2 at .0954 atm. (Note: these are the values to use for Kp) Using P1/n1 = P2/n2 forNOBr to determine the moles at equilibrium: .4926 atm/ .02 mole = .3018 atm/ n2 n2 = .0123 molesNOBr [NOBr] = .0123 moles/1 L = .0123M We can use the same equation as before to find the moles of NO: .4926 atm/ .02 mole = .1908 atm / n2 n2 = .0077 molesNO [NO] = .0077 mole/ 1L = .00775M (we can use the same P1 and n1 values becausethe same moles of gas always exert the same amount of pressure whentemperature and volume are the same) Br2 has half the pressure of NO, and therefore half the moles, andtherefor half the molarity: [Br2] = .00388M Kc = (.00775)2(.00388)/ (.0123)2 =.00154 3. Kp = .19082 x .0954 /.30182 = .0381 4. Because the temperature has not changed, we can use theequilibrium constants we already calculated. [NOBr] = .2 mole/ 2L = .1M [NO] = .15 mole / 2L = .075M [Br2] = .30 mol / 2L = .15M Qc = .0752 x .15 / .12 = .0843 Because Qc is greater than Kc, the reaction must shift left inorder to reduce the concentration of the products and increase theconcentration of the reactants. NOBr will be produced.