Aluminum metal and bromine liquid (red) react violently to make aluminum bromide

Question

Aluminum metal and bromine liquid (red) react violently to make aluminum bromide (white powder). One way to represent this equilibrium is:

Al(s) + 3/2 Br2(l)AlBr3(s)

We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.

1) 2 AlBr3(s) 2 Al(s) + 3 Br2(l) K1 =

2) AlBr3(s) Al(s) + 3/2 Br2(l) K2 =

3) 2 Al(s) + 3 Br2(l) 2 AlBr3(s) K3 =

Explanation / Answer

1) 2 AlBr3(s) 2 Al(s) + 3 Br2(l)

this is inverted,

and has the factor is 2x, that is ^2

K1 =^-1

overall factor:

K1 = K^-2

2)

2) AlBr3(s) Al(s) + 3/2 Br2(l) K2 =

this is inverted, therefore ^1

K2 = K^-1

3)

3) 2 Al(s) + 3 Br2(l) 2 AlBr3(s) K3 =

this is 2x factor so

K3 = K^2

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