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jolli1 [7]
2 years ago
10

At 580 k, 6. 42% of the molecules are in the chair form. calculate the value of the equilibrium constant for the reaction as wri

tten above.
Chemistry
1 answer:
Ann [662]2 years ago
5 0
Equilibrium chemical reaction between chair and boat forms is:
Kc = \frac{[Boat form]}{[Chair form]}
Suppose the total number of molecules is x 
Number of molecules of chair form is 6.42/100 x
Number of molecules of boat form is 93.58/100 x
Kc = \frac{(0.9358x)}{(0.0642x)} = 14.6
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For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow
lora16 [44]

Answer:

Rate constant of the reaction is 3.3\times 10^{-3} M^{-2} s^{-1}.

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Expression of rate law of the reaction will be written as:

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Rate(R) of the reaction in trail 1 ,when :

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R=9.0\times 10^{-5} M/s

9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c...[1]

Rate(R) of the reaction in trail 2 ,when :

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Rate(R) of the reaction in trail 3 ,when :

[A]=0.60 M,[B]=0.30 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

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Rate(R) of the reaction in trail 4 ,when :

[A]=0.60 M,[B]=0.60 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

R=k[A]^2[B]^0[C]^1

Rate constant of the reaction = k

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k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}

k=3.3\times 10^{-3} M^{-2} s^{-1}

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