Part a).
Kp = Kc*(RT)n
i.e. 4.7*107 = Kc * {0.08206*(150+273.15)}1-2
Therefore, Kc = 2.42*1011
3. The equilibrium constant K, for the formation of COBra, is given below. Kp = 4.7...
The equilibrium constant Kp for the formation of COBr2, is given below. CO(g) + Br2(g) <-----> COBr2(g) Kp = 4.7 x 107 at 150°C (a) Calculate Kc for the reaction as written above. (1 pt) Answer ___________________ (b) Calculate Kp for the following reaction at 150°C. (Refer to the reaction and Kp = 4.7 x 107 from part a.) (1 pt) 3COBr2(g) <------> 3CO(g) + 3Br2(g) Answer ___________________ (c) At a HIGHER temperature, a mixture of CO and Br2 is...
Consider the equilibrium between COBr2, CO and Br2 COBr2(g) at 382 K CO(g) + Brz(9) K= 2.08 The reaction is allowed to reach equilibrium in a 13.8-L flask. At equilibrium, [COBr2] = 4.43x10-2 M, [CO] = 0.304 M and [Br2] = 0.304 M. (a) The equilibrium mixture is transferred to a 6.90-L flask. In which direction will the reaction proceed to reach equilibrium? (b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to a 6.90-L...
Consider the equilibrium between
COBr2, CO and
Br2.
COBr2(g) CO(g)
+ Br2(g) K = 0.254 at
350 K
The reaction is allowed to reach equilibrium in a
6.40-L flask. At equilibrium,
[COBr2] = 0.294 M,
[CO] = 0.274 M and
[Br2] = 0.274 M.
(a) The equilibrium mixture is transferred to a
12.8-L flask. In which direction will the reaction
proceed to reach equilibrium?
(b) Calculate the new equilibrium concentrations that result when
the equilibrium mixture is transferred to a...
Consider the equilibrium between COBr2, CO and Br2. COBr2(g) -->CO(g) + Br2(g) K = 1.84 at 380 K The reaction is allowed to reach equilibrium in a 13.8-L flask. At equilibrium, [COBr2] = 3.02×10-2 M, [CO] = 0.236 M and [Br2] = 0.236 M. (a) The equilibrium mixture is transferred to a 6.90-L flask. In which direction will the reaction proceed to reach equilibrium? _________to the right? to the left? (b) Calculate the new equilibrium concentrations that result when the...
Tutored Practice Problem 15.4.2 COUNTS TOWARDS GRADE Close Problem Predict and calculate the effect of volume change on an equilibrium system. Consider the equilibrium between COBr2, CO and Br2 COBr2(g) CO(g)+ Br2(g) K 0.338 at 354 K The reaction is allowed to reach equilibrium in a 5.30-L flask. At equilibrium, [COBr2] 0.295 M and [Br2] = 0.257 M, [CO] 0.295 M. (a) The equilibrium mixture is transferred to a 10.6-L flask. In which direction will the reaction proceed to reach...
Close Problem Tutored Practice Problem 15.4.2 COUNTS TOWARDS GRADE Predict and calculate the effect of volume change on an equilibrium system. Consider the equilibrium between COBr2, CO and Br2. COBr2(g) PCO(g) + Brz(9) K = 0.254 at 350 K The reaction is allowed to reach equilibrium in a 8.60-L flask. At equilibrium, [COBr2] = 0.288 M, [CO] = 0.270 M and [Brz] = 0.270 M. (a) The equilibrium mixture is transferred to a 17.2-L flask. In which direction will the...
6) Distrubuting Equilibrium a) The equilibrium constant, K, for the following reaction is 10.5 at 350 K 2CH2Cl2(g)CH4(g + CCl4(g) An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.32x10-2 M CH2C2, 0.172 M CH4 and 0.172 M CCI4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 9.58x102 mol of CH4(g is added to the flask? CH2Cl2l [CH4 [CCI4] b) The equilibrium constant, K, for the...
Carbonyl bromide decomposes to carbon monoxide and bromine. COBr2(g) = CO(g) + Br2(g) Kc is 0.190 at 73 °C. If you place 0.514 mol of COBr2 in a 1.00-L flask and heat it to 73 °C, what are the equilibrium concentrations of COBr2, CO, and Br? [COBr2] = mol/L [CO] = mol/L [Bry] = mol/L The equilibrium constant for the dissociation of iodine molecules to iodine atoms 12(g) = 2 (g) is 3.76 x 10-3 at 1000 K. Suppose 0.338...
The reaction below has an equilibrium constant of Kp = 2.26 x 104 at 298 K. CO(g) + 2H2(g) = CH3 OH(9) Predict whether reactants or products will be favored at equilibrium in each reaction. Drag the appropriate items to their respective bins. Reset Help Kp = 150 Kp = 1.96 x 10-9 K, = 4.42 x 10 Reactants favored Products favored Submit Request Answer
Consider the reaction: 2NO(g)+Br2(g)⇌2NOBr(g) Kp=28.4 at 298 K In a reaction mixture at equilibrium, the partial pressure of NO is 101 torr and that of Br2 is 150 torr . What is the partial pressure of NOBr in this mixture? Express your answer using three significant figures.