The equilibrium constant for the following reaction is 50.5 at 448 oC.
H2(g) + I2(g) ⇌ 2 HI(g)
a. Find DGorxn at 448 oC. (This is the change in free energy when the partial pressure of each gas is 1 atm and the T is 448 oC.)
b. What is DGrxn when T = 448 oC and PH2 = 1 atm, PI2 = 0.1 atm, PHI = 0.1 atm?
c. Is the forward reaction spontaneous for the conditions listed in part b? Explain!
The equilibrium constant for the following reaction is 50.5 at 448 oC. H2(g) + ...
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,...
At 49 oC the equilibrium constant for the reaction: 2 HI(g) H2(g) + I2(g) is KP = 4.83e-11. If the initial pressure of HI is 0.00862 atm, what are the equilibrium partial pressures of HI, H2, and I2? p(HI) = p(H2) = . p(I2) = .
Consider this reaction at 298 K: H2 (g) + I2 (g) ⇌ 2 HI(g) Calculate ΔGrxn under the following conditions: PH2 (g) = 0.161 atm PI2 (g) = 0.186 atm PHI(g) = 0.307 atm
A,) ΔG o for the reaction H2(g) + I2(g) ⇌ 2HI(g) is 2.60 kJ/mol at 25°C. Calculate ΔG, and predict the direction in which the reaction is spontaneous. The initial pressures are: PH2 = 3.10 atm PI2 = 1.5 atm PHI 1.75 atm ΔG = kJ/mol b.)The reaction is spontaneous in the forward direction. The reaction is spontaneous in the reverse direction. Cannot be determined.
. Consider the reaction: H2(g) + I2(g) ? 2HI(g) A reaction mixture at equilibrium at 175 K contains PH2 = 0.958 , PI2 = 0.877 , and PHI= 0.020 . A second mixture not equilibrium, but at the same temperature,contains PH2 = PI2 = 0.621 , and PHI = 0.101 . a) For the second mixture, calculate Q. b) Determine whether the reaction in the second mixture will shift right or shift left.c) Construct an ICE table. Keep in mind your...
The reaction below was observed in a 4.0 L flask at 445 oC. H2 (g) + I2 (g) ↔ 2 HI (g) Initially, 0.20 atm of each gas (H2, I2 and HI) was placed in the flask. If the kp = 50.2 atm at 445oC, what is the equilibrium pressure (in atm) of HI (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...
H2(g) + I2(g) ⇌ 2HI(g) ΔG o for the reaction is 2.60 kJ/mol at 25°C. What is the minimum partial pressure of I2 required for the reaction to be spontaneous in the forward direction at 25°C if the partial pressures of H2 and HI are 4.1 and 2.05 atm, respectively?
Enter your answer in the provided box. Given the reaction: H2(g) + I2(g) ⇌ 2HI(g) ΔG o for the reaction is 2.60 kJ/mol at 25degree C. What is the minimum partial pressure of I2 required for the reaction to be spontaneous in the forward direction at 2degrees C if the partial pressures of H2 and HI are 3.5 and 1.75 atm, respectively?
When HI(g) is heated to 700 K, it reversibly decomposes to H2(g) and I2(g). The reaction is- 2 HI(g) ⇌ H2(g) + I2(g). A 15.00-L vessel at 700 K initially contains HI(g) at a pressure of 4.00 atm. When equilibrium is reached, it is found that the partial pressure of H2(g) is 0.387 atm. What is the partial pressure of HI(g) at equilibrium? A) 4.00 atm B) 3.61 atm C) 3.23 atm D) 4.39 atm E) 0.387 atm