Question

at a certain temperature, the equilibrium constant ( k ) for the following reaction is ( 0.75 ):
( mathrm{br}_{2}(mathrm{~g})+mathrm{ocl}_{2}(mathrm{~g})
ightleftharpoons mathrm{brocl}(mathrm{g})+mathrm{brcl}(mathrm{g}) )
use this information to complete the following table.

suppose a ( 23. mathrm{~l} ) reaction vessel is filled with ( 0.59 mathrm{~mol} ) of ( mathrm{brocl} ) and ( 0.59 mathrm{~mol} ) of ( mathrm{brcl} ). what can you say about the composition of the mixture in the vessel at equilibrium?

( \bigcirc ) there will be very little ( mathrm{br}_{2} ) and ( mathrm{ocl}_{2} ).<br>( \bigcirc ) there will be very little ( mathrm{brocl} ) and ( mathrm{brcl} ).<br>( \bigcirc ) neither of the above is true.

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what is the equilibrium constant for the following reaction?<br>round your answer to 2 significant digits.<br>( mathrm{brocl}(mathrm{g})+mathrm{brcl}(mathrm{g})

ightleftharpoons mathrm{br}_{2}(mathrm{~g})+mathrm{ocl}_{2}(mathrm{~g}) ) | ( k=square ) |<br>| what is the equilibrium constant for the following reaction?<br>round your answer to 2 significant digits.<br>( 2 mathrm{br}_{2}(mathrm{~g})+2 mathrm{ocl}_{2}(mathrm{~g})
ightleftharpoons 2 mathrm{brocl}(mathrm{g})+2 mathrm{brcl}(mathrm{g}) ) | ( k=square ) |

Explanation:

Step1: Calculate the reaction quotient \(Q\)

The formula for \(Q\) is \(Q=\frac{[BrOCl][BrCl]}{[Br_{2}][OCl_{2}]}\). Given \(n(BrOCl) = 0.59\space mol\), \(n(BrCl)=0.59\space mol\), and \(V = 23\space L\). So \([BrOCl]=\frac{0.59}{23}\space M\), \([BrCl]=\frac{0.59}{23}\space M\), and \([Br_{2}]=[OCl_{2}]=0\space M\) (initially). Then \(Q=\infty\) (since denominator is \(0\)). Since \(Q>K = 0.75\), the reaction will shift to the left. So there will be very little \(BrOCl\) and \(BrCl\) at equilibrium.

Step2: Find \(K\) for \(BrOCl(g)+BrCl(g)

ightleftharpoons Br_{2}(g)+OCl_{2}(g)\)
If the original reaction is \(Br_{2}(g)+OCl_{2}(g)
ightleftharpoons BrOCl(g)+BrCl(g)\) with \(K_1 = 0.75\), then for the reverse reaction \(BrOCl(g)+BrCl(g)
ightleftharpoons Br_{2}(g)+OCl_{2}(g)\), \(K_2=\frac{1}{K_1}\). So \(K_2=\frac{1}{0.75}\approx1.3\)

Step3: Find \(K\) for \(2Br_{2}(g)+2OCl_{2}(g)

ightleftharpoons 2BrOCl(g)+2BrCl(g)\)
If the original reaction is \(Br_{2}(g)+OCl_{2}(g)
ightleftharpoons BrOCl(g)+BrCl(g)\) with \(K_1 = 0.75\), for the reaction \(aA + bB
ightleftharpoons cC + dD\), if we multiply the reaction by \(n\), \(K'=(K)^n\). Here \(n = 2\), so \(K=(0.75)^2=0.56\)

Answer:

• There will be very little \(BrOCl\) and \(BrCl\).
• \(K = 1.3\)
• \(K = 0.56\)