_____ The k_eq values apply to the following two equations at 25 degree C: H_2(g
ID: 484303 • Letter: #
Question
_____ The k_eq values apply to the following two equations at 25 degree C: H_2(g) + l_2(g) 2HI(g) K_eq= 6.17 times 10^2 H_2(g) + Br_2(g) 2HBr(g) K_eq = 1.88 times 10^19 Calculate K_eq for the equation: l_2(g) + 2HBr(g) - Br_2(g) + 2HI(g) 3.28 rightarrow10^17. 3.04 times 10^16. 1.56 times 1022. 8.64 times 10^21. 4.29 times 10^17. 8.05 times 10^21. 1.00. 1.16 times 10^22. _____ K_eq = 1.03 times 10^5 for the reaction CO(g) + H_2O(g) 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.. 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) + Q_2(g) CO_2(g), if the temperature is increased, the reaction shifts to the left. shifts to the right. does not change. _____ For the reaction C(s) + CO_2(g) 2CO(g), at equilibrium at 850 degree C the relative molar amount of C, and 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. 1.70. 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 it increases if the concentrations are decreased.Explanation / Answer
you can do it by simply putting values of Keq for the three reactions ...
you see ...Keq (1) = [HI]2/ [H2][I2]
for second equation ....Keq (2) = [HBr]2/ [H2][Br2]
and for final equation ...
Keq(3) = [Br2][HI]2/ [HBr]2[I2]
by taking ratio of Keq(1) and Keq(2)...
we have ...Keq (3) ....
as [HI]2/ [H2][I2]/ [HBr]2/ [H2][Br2] = [Br2][HI]2/ [HBr]2[I2]
hence their numerical value = 6.17 X 102/ 1.88 X 1019
= 3.28 X 10-17
2) we have Keq ...and we also have the concentrations of the rest species ...hence all we have to do is to put the value in the Keq ..as ..
Keq = [CO2][H2]/[CO][H2O]
so just substituting value of the rest ...we have
1.03X 105 = 0.01 X [H2] / 0.002 X 0.0015
or [H2]= 1.03X 105 X 0.002 X 0.0015 /0.01
= 30.9 ..option D
3) according to le'chatlier's principle ...the equiblirium tries to shift in a direction to undo the change ....or since this is a exothermic reaction(heat is evolved)....and if we increase the temperature ..then the equiblirium has to shift in opposite direction.....hence shifts to left....option B
4) C (s)+ CO2(g) ----> 2CO(g)
or Keq = [CO]2/[CO2] ......note that while calculating equlibrium constants ...we don't consider the conc of solids....
= 0.542 / 0.062
4.7 ..option D
5) a catalyst helps to achieve the reaction faster ...by providing an alternate mechanism ...it doesn't affect the equiblrium constant....so option ..A
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.