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3 years ago

The equilibrium constant for A + 2B → 3C is 2.1 * 10^-6

Chemistry

1 answer:

Mariana [72]3 years ago

7 0

Answer:

b- 4.4 * 10^-12.

Explanation:

Hello.

In this case, as the reaction:

A + 2B → 3C

Has an equilibrium expression of:

$K_1=\frac{[C]^3}{[A][B]^2}=2.1x10^{-6}$

If we analyze the reaction:

2A + 4B → 6C

Which is twice the initial one, the equilibrium expression is:

$K_2=\frac{[C]^6}{[A]^2[B]^4}$

It means that the equilibrium constant of the second reaction is equal to the equilibrium constant of the first reaction powered to second power:

$K_2=K_1^2$

Thus, the equilibrium constant of the second reaction turns out:

$K_2=(2.1 * 10^{-6})^2\\\\K_2=4.4x10^{-12}$

Therefore, the answer is b- 4.4 * 10^-12.

Best regards.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a) Splitting pattern for Ha= 2+1 , Triplet

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Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

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b) Splitting pattern for Ha= 2+1 , Triplet

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Explanation:

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For Proton Ha, number of non equivalent protons n= 2

Splitting pattern for Ha= 2+1 , Triplet

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Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

Considering the second question for Molecule 2 as shown on the third uploaded image

For Proton Ha, number of non equivalent protons n= 1

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For Proton Hb, number of non equivalent protons n= 3

Splitting pattern for Hb= 3+1=4 , Quartet

For Proton Hc, number of non equivalent protons n= 3

Splitting pattern for Hc= 3+1=4 , Quartet

For Proton Hd, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

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The IUPAC name is Butan-2-ol

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