1). The reaction N 2 ( g ) + 3 H 2 ( g ) 2 NH 3 ( g ) has a K p value of 1.5×10
ID: 548979 • Letter: 1
Question
1). The reaction N2(g) + 3 H2(g) 2 NH3(g) has a Kp value of 1.5×10–5. What is the Kc for this reaction at 500. °C?
7.3×101
9.0×109
6.0×102
4.0×109
2.5×102
2). The following equilibrium constant data is collected at 25°C:
1) N2(g) + O2(g) 2 NO(g) Kc = 4.8×10-31
2) 2 NOBr(g) 2 NO(g) + Br2(g) Kc = 0.50
What is the value of Kc for the following reaction at 25°C?
2 NOBr(g) N2(g) + O2(g) + Br2(g) Kc = ?
2.4×1031
3.7×1015
9.6×1031
1.0×1030
4.8
3). The equilibrium constant, Kp , for the reaction H2(g) + I2(g) 2 HI(g) is 55.2 at 425°C. A flask at 425°C contains 0.127 atm of hydrogen, 0.134 atm of iodine, and 1.055 atm of hydrogen iodide. Is the system at equilibrium?
Yes, the system is at equilibrium.
No, the reverse reaction must proceed to establish equilibrium.
No, the forward reaction must proceed to establish equilibrium.
Need to know the starting concentrations of all substances before deciding.
4). At 700 K, Kc = 1.56×10–2 for the reaction 2 HBr(g) H2(g) + Br2(g). In a given experiment, 0.050 mol of H2, and 0.050 mol Br2 are introduced into a 5.0-L flask. What is the equilibrium concentration of HBr? (Round answer to 2 sig figs and don't use scientific notation)
5). The reaction system of POCl3(g) POCl(g) + Cl2(g) is at equilibrium in a flask. Which of the following statements describes the behavior of the system if some of the chlorine gas is removed from the reaction flask?
The partial pressure of POCl will decrease while the partial pressure of Cl2 increases as equilibrium is established.
Chlorine will be consumed as equilibrium is established.
POCl will be consumed as equilibrium is established.
POCl3 will be consumed as equilibrium is established.
The volume will have to decrease before equilibrium can be reestablished.
6). Which of the following will increase the pressure of N(g) according to the following reaction?
2 N(g) N2(g)
Remove N2
Decrease the temperature
Add N
Increase the temperature
7.3×101
9.0×109
6.0×102
4.0×109
2.5×102
2). The following equilibrium constant data is collected at 25°C:
1) N2(g) + O2(g) 2 NO(g) Kc = 4.8×10-31
2) 2 NOBr(g) 2 NO(g) + Br2(g) Kc = 0.50
What is the value of Kc for the following reaction at 25°C?
2 NOBr(g) N2(g) + O2(g) + Br2(g) Kc = ?
2.4×1031
3.7×1015
9.6×1031
1.0×1030
4.8
3). The equilibrium constant, Kp , for the reaction H2(g) + I2(g) 2 HI(g) is 55.2 at 425°C. A flask at 425°C contains 0.127 atm of hydrogen, 0.134 atm of iodine, and 1.055 atm of hydrogen iodide. Is the system at equilibrium?
Yes, the system is at equilibrium.
No, the reverse reaction must proceed to establish equilibrium.
No, the forward reaction must proceed to establish equilibrium.
Need to know the starting concentrations of all substances before deciding.
4). At 700 K, Kc = 1.56×10–2 for the reaction 2 HBr(g) H2(g) + Br2(g). In a given experiment, 0.050 mol of H2, and 0.050 mol Br2 are introduced into a 5.0-L flask. What is the equilibrium concentration of HBr? (Round answer to 2 sig figs and don't use scientific notation)
5). The reaction system of POCl3(g) POCl(g) + Cl2(g) is at equilibrium in a flask. Which of the following statements describes the behavior of the system if some of the chlorine gas is removed from the reaction flask?
The partial pressure of POCl will decrease while the partial pressure of Cl2 increases as equilibrium is established.
Chlorine will be consumed as equilibrium is established.
POCl will be consumed as equilibrium is established.
POCl3 will be consumed as equilibrium is established.
The volume will have to decrease before equilibrium can be reestablished.
6). Which of the following will increase the pressure of N(g) according to the following reaction?
2 N(g) N2(g)
Remove N2
Decrease the temperature
Add N
Increase the temperature
Explanation / Answer
1)
T= 500.0 oC
= (500.0+273) K
= 773 K
delta n = number of gaseous molecule in product - number of gaseous molecule in reactant
delta n = -2
Kp= Kc (RT)^deltan
1.5*10^-5 = Kc *(0.0821*773.0)^(-2)
Kc = 6.0*10^-2
Answer: 6.0*10^-2
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