0ne of the elementary steps is given below, at 956 degree C 2NaHCO_3(s) = Na_2CO

Question

0ne of the elementary steps is given below, at 956 degree C 2NaHCO_3(s) = Na_2CO_2(s) + H_2CO_3(g) K_1 = 1.04 times 10^2 K_2 = 1.5 times 10^-1 2NaHCO_2(s) Na_2CO_3(s) + CO_2(g) + H_2(g) What must be the other reaction which when added, would give the net reaction? Think of the intermediate thing we already did in chapter 13 Identify the intermediate in the mechanism above, explaining your choice Write the equilibrium constant expression for steps 1) and 2). Show how you can get the equilibrium constant expression for the net equation from the two steps See text p 636. Calculate the equilibrium constant for the net equation from the elementary steps

Explanation / Answer

intermediate reaction is

H2CO3----->H2O+CO2

Intermediate is H2CO3

Net equilibrium constant is Ka=Ka1+Ka2

Ka=0.0104+0.15

Ka=0.1604

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