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In: Chemistry

Given the following reactions: COBr2 (g) --------------- > CO (g) + Br2 (g)              K = .190...

Given the following reactions:

COBr2 (g) --------------- > CO (g) + Br2 (g)              K = .190 at 73 °C

2 BrCl (g) ------------ > Br2 (g) + Cl2 (g)          K = 310 at 73 °C (estimated)

Find the equilibrium constant at 73 °C for: CO + BrCl ------- > COBr2 + Cl2

Solutions

Expert Solution

Find the equilibrium constant at 73 °C for: CO + BrCl ------- > COBr2 + Cl2

COBr2 (g) --------------- > CO (g) + Br2 (g)              K = .190 at 73 °C---1

2 BrCl (g) ------------ > Br2 (g) + Cl2 (g)          K = 310 at 73 °C---2

CO + BrCl ------- > COBr2 + Cl2

To find the above reaction reverse the eq 1.

CO (g) + Br2 (g)      --------------- >  COBr2 (g)   K’= 1/K=1/0.190=5.36 ---3

On reversing the equation new equilibrium constant is 1/K    

Now add eq 2 and 3

CO (g) + Br2 (g)      --------------- >  COBr2 (g)

2 BrCl (g) ------------ > Br2 (g) + Cl2 (g)             

CO + 2BrCl ------- > COBr2 + Cl2

To add equation equilibrium constant are multiplied

K = 5.36 *310

K= 1661.6

queen_honey_blossom answered 2 years ago
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