283.3 for the equilibrium reaction below at 131.8 °C, find Ke If Kp = 2H2(g)2g)=2H20()
If Kp = 307.7 for the equilibrium reaction below at 326.2 °C, find Ko CH_(9) +202(g) = CO2(g) +2H20(1)
The reaction below has an equilibrium constant of Kp=2.26×104 at 298 K. CO(g)+2H2(g)⇌CH3OH(g) Part A Calculate Kp for the reaction below. CH3OH(g) CO(g)+2H2(g) Submit My Answers Give Up Part B Reactants will be favored at equilibrium. O Products will be favored at equilibrium. Submit My Answers Give Up Part C Calculate Kp for the reaction below. 를 CO (g) + H2 (g)- CH, OH (g) K=
QUESTION 4 The following exothermic reaction is allowed to reach equilibrium. 2H2(g) + O2(g) = 2H20(g) What happens to the reaction when the temperature is decreased? o Favors forward reaction Favors reverse reaction The reaction is still at equilibrium. 2 points Saved
The equilibrium constant Kp for the reaction 2H2O(g)--> 2H2(g)+O2(g) is 2x10^-42 at 25 degrees C. (a) what is Kc for the reaction at the same temperature? (b) the very small value of Kp (and Kc) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.
The relevant reaction is 10- Calculate Ke for this equilibrium at 25°C. Kp for the reaction of SO(g) with O, to produce SOS) is 3 250,(g) + O2(g) + 250,)) O A 3x1024 OB.5 x 1021 OC 1020 OD.7 x 1025 O Al me e
73. Consider the reaction: CO(g) + 2H2(g) <--> CH3OH(g) Kp= 2.26X10E4 at 25 C Calculate ΔGorxnfor the reaction at 25C under each of the following conditions: a. standard conditions b. at equilibrium c. PCH3OH= 1.0 atm; PCO= PH2 = 0.010 atm
At a certain temperature, Kp = 4.5 x 103 for the reaction: 2H2S(g) ⇌ 2H2(g) + S2(g) If P[H2S] = 0.0050 atm, P[H2] = 0.50 atm, and P[S2] = 0.75 atm, then answer choices: Qp < Kp and the reaction will shift toward products. Qp < Kp and the reaction will shift toward reactants Qp > Kp and the reaction will shift toward products. Qp = Kp and the reaction is at equilibrium. Qp > Kp and the reaction...
8. If the reaction 2H2S(g) = 2H2(g) + S2(g) is carried out at 1065°C, Kp = 0.0120. Starting from pure H2S introduced into an evacuated vessel at 1065°C, what will the total pressure in the vessel at equilibrium if the equilibrated mixture contains 0.300 atm of H2(g)? Answer 1.51 atm (1.5 pts)
At a certain temperature, Kp = 4.5 x 103 for the reaction: 2H2S(g) = 2H2(g) + S2(g) If P[H2S] = 0.0050 atm, P[H2] = 0.50 atm, and P[S2] = 0.75 atm, then Qp > Kp and the reaction will shift toward products. O Qp > Kp and the reaction will shift toward reactants. O Qp <Kp and the reaction will shift toward products. O Qp = Kp and the reaction is at equilibrium. Qp <Kp and the reaction will shift...
For the reaction: 2H20(g) = 2H2(g) + O2(g) Kc = 2.40 x 10-3 at a given temperature. At equilibrium, it is found that [H20] = 0.290 M and [H2] = 1.80 x 10-2 M. If the reaction is run in a 8.50 L container, how many moles of O2 are present at equilibrium? moles O2 Tries 0/99 Submit Answer