The equilibrium constant (K_c) for the formation of nitrosyl chloride, an orange
ID: 937449 • Letter: T
Question
The equilibrium constant (K_c) for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine is 3 Times 10^7 at a certain temperature. 2 NO(g) + Cl_2(g) rightarrow 2NOCl(g) In an experiment, 1.7 Times 10^-2 mole of NO, 5.1 Times 10^-3 mole of Cl_2, and 9.5 moles of NOC1 are mixed in a 2.9 L flask. What is Qc for the experiment? In which direction will the system proceed to reach equilibrium? For the synthesis of ammonia N_2(g) + 3H_2(g) 2NH_3(g) the equilibrium constant K_c at 375 degree C is 1.2. Starting with [H_2]_0 = 0.76 m, [N_2]_0 = 0.60 M, and [NH_3]_0 = 0.48 M, which gases will have increased in concentration and which will have decreased in concentration when the mixture comes to equilibrium? At 1000 K, a sample of pure NO_2 gas decomposes: The equilibrium constant k_p is 158. Analysis shows that the partial pressure of O_2 is 0.25 atm at equilibrium. Calculate the pressure of NO and NO_2 in the mixture.Explanation / Answer
[N0] = 0.017/ 2.9 = 0.00586 moles
[Cl2]= 0.0051/2.9 = 0.00176 moles
[NOCl] = 9.5 / 2.9 = 3.27 moles
as we know that Qc = [NOCl]2 /[N0]2 [Cl2] = (3.27)2 / (0.00586)2 (0.00176) = 1.769 *108
As Kc = 3*107 and we know if Kc < Qc then the reaction will proceed backward to attain equillibrium .
2) using Qc = [NH3]2 / [H2]3 [N2]
= (0.48)2 /(0.76)3 * (0.60) = 0.2304 / 0.3465 = 0.6649
As Kc = 1.2 and we know if Kc > Qc then the reaction will proceed forward to attain equillibrium .
the concerntration of [N2] and [H2] will decrease and that of [NH3] is incresed when systemcomes to equillibrium .
3)
Kp = P²(NO) P(O2) / P²(NO2)
158 = [0.50]^2 [0.25] / (NO2) 2
(NO2) 2= [0.50]^2 [0.25] / 158 = 0.0003956
(NO2) = 0.01989 = 0.02 atm
as 2NO2(g) --> 2NO(g) + O2(g)
when O2 is 0.25 atm is produced, twice as many moles of NO are produced = 0.50 atm for NO
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