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1. At a given temperature the Kp = 4.76x10-4 for the reaction: H2(g) + 12(g) =...
The equilibrium constant, Kp, for the following reaction is 1.80x10-2 at 698K. 2HI(g) =H2(g) +12(g) If...
The equilibrium constant, Kp, for the following reaction is 1.80x10-2 at 698K. 2HI(g) =H2(g) +12(g) If an equilibrium mixture of the three gases in a 14.3 L container at 698K contains HI at a pressure of 1.09 atm and H, at a pressure of 0.929 atm, the equilibrium partial pressure of I, is atm. Submit Answer Retry Entire Group 9 more group attempts remaining
(1). The equilibrium constant, Kp, for the following reaction is 1.80×10-2 at 698K. 2HI(g) =H2(g) +...
(1). The equilibrium constant, Kp, for the following reaction is 1.80×10-2 at 698K. 2HI(g) =H2(g) + I2(g) If an equilibrium mixture of the three gases in a 15.5 L container at 698K contains HI at a pressure of 0.399 atm and H2 at a pressure of 0.562 atm, the equilibrium partial pressure of I2 is atm. (2). Consider the following reaction: PCl5(g) =PCl3(g) + Cl2(g) If 1.17×10-3 moles of PCl5, 0.217 moles of PCl3, and 0.351 moles of Cl2 are at...
Given the reaction at a certain temperature: H2(g) + I2(g) ↔ 2 HI(g). At equilibrium, the...
Given the reaction at a certain temperature: H2(g) + I2(g) ↔ 2 HI(g). At equilibrium, the partial pressure of HI is 3.8×10–3 atm and the partial pressures of H2 and I2 are 0.20 atm each. The Kp of this reaction is [X]. (Fill in the blank; report with correct number of significant figures.
At 6 oC the equilibrium constant for the reaction: 2 HI(g) H2(g) + I2(g) is KP...
At 6 oC the equilibrium constant for the
reaction:
2 HI(g) H2(g) + I2(g)
is KP = 2.66e-11. If the initial pressure of HI is
0.00837 atm, what are the equilibrium partial pressures of HI,
H2, and I2?
We were unable to transcribe this imageAt 6 °C the equilibrium constant for the reaction: 2 HI(g) = H2(g) + 12(g) is Kp = 2.66e-11. If the initial pressure of HI is 0.00837 atm, what are the equilibrium partial pressures of HI,...
A student ran the following reaction in the laboratory at 690. K: H2(g) +12(g) 22HI(g) When...
A student ran the following reaction in the laboratory at 690. K: H2(g) +12(g) 22HI(g) When he introduced H2(g) and 12(g) into a 1.00 L evacuated container, so that the initial partial pressure of H2 was 4.40 atm and the initial partial pressure of I was 3.28 atm, he found that the equilibrium partial pressure of HI was 5.78 atm. Calculate the equilibrium constant, Kp, he obtained for this reaction. Kp = Submit Answer Retry Entire Group 9 more group...
For the reaction H2 S(g) + I2 (s) ⇌ S(s) + 2 HI(g) Kp = 1.33×10–5...
For the reaction H2 S(g) + I2 (s) ⇌ S(s) + 2 HI(g) Kp = 1.33×10–5 at 333 K. What will be the total pressure of the gases above an equilibrium mixture if, at equilibrium, PHI = 0.010 × PH2 S? The answer is 0.134 atm. Please include work and explanation! Thank you :-)
a) H2(g) + 12 (g) = 2 HI(g) (3 points) A reaction mixture at equilibrium at...
a) H2(g) + 12 (g) = 2 HI(g) (3 points) A reaction mixture at equilibrium at 175 K contains Pue = 0.958 am, Pe=0.877 atm, and Pu=0.020 atm. A second reaction mixture, also at 175 K. contains Pie Pe=0.621 atm, and Pm 0.101 atm. Is the second reaction mixture at equilibrium? (Show all work for full credit) a IT b) (7 points) If not at equilibrium, what will be the partial pressure of HI when the reaction reaches equilibrium at...
Consider the reaction: CO(g) + H2O(g) -><- CO2(g) + H2(g) Kp = 0.0871 at 1000 K...
Consider the reaction: CO(g) + H2O(g) -><- CO2(g) +
H2(g)
Kp = 0.0871 at 1000 K
A reaction mixture originally contains a CO partial pressure
of 1744 torr and a H2O partial pressure of 766 torr at 1000 K.
Caluculate the equilibrium partial pressures of each of the
products
6) (10 points) Consider the reaction: CO(g) + H2O(g) = CO2(g) + H2(g) Kp = 0.0871 at 1000 K A reaction mixture initially contains a CO partial pressure of 1 744...
4. Consider the reaction: 2 NO(g) + Br2 (g) 2 NOB (g) Kp = 31.8 at...
4. Consider the reaction: 2 NO(g) + Br2 (g) 2 NOB (g) Kp = 31.8 at 302 K a) If the initial partial pressures are Pro = 108 torr, Per; = 126 torr, ProBr = 275 torr, find Qp and determine which direction the reaction proceeds in to reach equilibrium. 139.17 b) When a mixture of NO, Bra, and H, reaches equilibrium at 302 K, Pno=133 torr and Pers 151 torr. Find PNobr (in torr). [335 torr] 5. lodine gas...
Consider the following reaction: CO(g)+H2O(g)⇌CO2(g)+H2(g) Kp=0.0611 at 2000 K A reaction mixture initially contains a CO...
Consider the following reaction: CO(g)+H2O(g)⇌CO2(g)+H2(g) Kp=0.0611 at 2000 K A reaction mixture initially contains a CO partial pressure of 1346 torr and a H2O partial pressure of 1762 torr at 2000 K. A.) Calculate the equilibrium partial pressure of CO2. B.) Calculate the equilibrium partial pressure of H2.
The equilibrium constant, Kp, for the following reaction is 1.80x10-2 at 698K. 2HI(g) =H2(g) +12(g) If an equilibrium mixture of the three gases in a 14.3 L container at 698K contains HI at a pressure of 1.09 atm and H, at a pressure of 0.929 atm, the equilibrium partial pressure of I, is atm. Submit Answer Retry Entire Group 9 more group attempts remaining
At 6 oC the equilibrium constant for the
reaction:
2 HI(g) H2(g) + I2(g)
is KP = 2.66e-11. If the initial pressure of HI is
0.00837 atm, what are the equilibrium partial pressures of HI,
H2, and I2?
We were unable to transcribe this imageAt 6 °C the equilibrium constant for the reaction: 2 HI(g) = H2(g) + 12(g) is Kp = 2.66e-11. If the initial pressure of HI is 0.00837 atm, what are the equilibrium partial pressures of HI,...
A student ran the following reaction in the laboratory at 690. K: H2(g) +12(g) 22HI(g) When he introduced H2(g) and 12(g) into a 1.00 L evacuated container, so that the initial partial pressure of H2 was 4.40 atm and the initial partial pressure of I was 3.28 atm, he found that the equilibrium partial pressure of HI was 5.78 atm. Calculate the equilibrium constant, Kp, he obtained for this reaction. Kp = Submit Answer Retry Entire Group 9 more group...
a) H2(g) + 12 (g) = 2 HI(g) (3 points) A reaction mixture at equilibrium at 175 K contains Pue = 0.958 am, Pe=0.877 atm, and Pu=0.020 atm. A second reaction mixture, also at 175 K. contains Pie Pe=0.621 atm, and Pm 0.101 atm. Is the second reaction mixture at equilibrium? (Show all work for full credit) a IT b) (7 points) If not at equilibrium, what will be the partial pressure of HI when the reaction reaches equilibrium at...
Consider the reaction: CO(g) + H2O(g) -><- CO2(g) +
H2(g)
Kp = 0.0871 at 1000 K
A reaction mixture originally contains a CO partial pressure
of 1744 torr and a H2O partial pressure of 766 torr at 1000 K.
Caluculate the equilibrium partial pressures of each of the
products
6) (10 points) Consider the reaction: CO(g) + H2O(g) = CO2(g) + H2(g) Kp = 0.0871 at 1000 K A reaction mixture initially contains a CO partial pressure of 1 744...
4. Consider the reaction: 2 NO(g) + Br2 (g) 2 NOB (g) Kp = 31.8 at 302 K a) If the initial partial pressures are Pro = 108 torr, Per; = 126 torr, ProBr = 275 torr, find Qp and determine which direction the reaction proceeds in to reach equilibrium. 139.17 b) When a mixture of NO, Bra, and H, reaches equilibrium at 302 K, Pno=133 torr and Pers 151 torr. Find PNobr (in torr). [335 torr] 5. lodine gas...
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