calculate the equilibrium constant (kc) for the reaction at this temperature. co
ID: 487220 • Letter: C
Question
calculate the equilibrium constant (kc) for the reaction at this temperature. co(g)+2h2(g)->ch3oh(g). a reaction mixture in a 5.16L flask at a certain temperature initially contains 27.3g co and 2.33g h2. at equilibrium the flask contains 8.64g ch3oh calculate the equilibrium constant (kc) for the reaction at this temperature. co(g)+2h2(g)->ch3oh(g). a reaction mixture in a 5.16L flask at a certain temperature initially contains 27.3g co and 2.33g h2. at equilibrium the flask contains 8.64g ch3oh calculate the equilibrium constant (kc) for the reaction at this temperature. co(g)+2h2(g)->ch3oh(g). a reaction mixture in a 5.16L flask at a certain temperature initially contains 27.3g co and 2.33g h2. at equilibrium the flask contains 8.64g ch3ohExplanation / Answer
calculate the equilibrium constant (kc) for the reaction at this temperature. co(g)+2h2(g)->ch3oh(g). a reaction mixture in a 5.16L flask at a certain temperature initially contains 27.3g co and 2.33g h2. at equilibrium the flask contains 8.64g ch3oh
Initial Moles of CO = Mass of CO / molecular mass = 27.3 / 28 = 0.975 moles
Initial concentration = Moles / volume = 0.975 / 5.16 = 0.189 M
Initial Moles of H2 = Mass / molar mass = 2.33 /2 = 1.165 moles
Initial concentration = 1.165 / 5.16 = 0.226 M
Moles of CH3OH at equilibrium = Mass / molar mass = 8.64 / 32 = 0.27 moles
CO(g) + 2H2(g) ---> CH3OH(g)
Initial moles 0.975 1.165 0
Change -x -2x +x
Equilibrium 0.975-x 1.165-2x x
x = 0.27 moles
Moles of CO = 0.975 - 0.27 = 0.705 . concentration of CO = Moles / volume = 0.705 / 5.16 = 0.137 M
Moles of H2O = 1.165- 2(0.27) = 0.625 , concentration of H2O =Moles / volume = 0.625/ 5.16 = 0.121 M
[CH3OH] = Moles / volume = 0.27 / 5.16 = 0.0523
Kc = [CH3OH] / [CO] [ H2]2 =0.0523 / (0.137(0.121)2 = 26.07
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