K_eq = 1.03 times 10^5 for the reaction CO (g) + H_2O(g) doubleheadarrow CO_2(g)
ID: 484314 • Letter: K
Question
K_eq = 1.03 times 10^5 for the reaction CO (g) + H_2O(g) doubleheadarrow CO_2(g) + H_2(g). A 1-liter flask at equilibrium has 0.002 mole of CO, 0.0015 mole of H_2O, and 0.01 mole of CO_2. How many moles of H_2 exist in the flask? 1.24. 0.39. 5.20. 30.9. 0.0034. 0.87. For the exothermic equilibrium CO (g) + O_2(g) doubleheadarrow CO_2(g), if the temperature is increased to reaction shifts to the left shifts to the right does not change. For the exothermic equilibrium C(s) + CO_2(g) doubleheadarrow 2CO(g), at equilibrium at 850 degree C the relative molar amounts of C, CO_2, and CO in a 1.00-L flask are 0.050, 0 062, and 0.54 respectively. what is the value of the equilibrium constant K? 14.25. 0.030. 0.15. 4.70. 2.82. 2.00. What effect does a catalyst have on the value of the K_eq for a chemical reaction? no effect. it increases. it decreases. it increases if the concentrations are increased it increases if the concentrations are decreased For the reaction2A + 2B doubleheadarrow C, K_eq = 5. What is K_eq for the reaction A + B = 1/2 AC? 50. 25. 100. 7.07. 2500. 1000. 5.0. 500.Explanation / Answer
3)Ans- " A" Equilibrium shifts to left
An equilibrium is if exothermic (i.e.- release heat) in forward direction then it must be endothermic(i.e.-heat absorbed) in backward direction. According to Lechatelier principle if an equilibrium is exothermic in nature then an increase in temperature shifts in direction where it is endothermic i.e.- towards left
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