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Zigmanuir [339]
2 years ago
7

The product of the reaction between the following substances was treated with h2o. Draw the final product.

Chemistry
1 answer:
Viefleur [7K]2 years ago
5 0

Answer:

See explanation and picture

Explanation:

Question is incomplete, however, I found this question on several sites and most of them, have the same substances, so I'm gonna do it with that example to give you a hint of how to do it with yours, in case, it's a different substance.

First, we have a ketone and another substance, with MgCl. This is tipically the Grignard reagent, and this is often used to produce a tertiary alcohol.

The mechanism and final product is the following (See attached picture).

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For the reaction 2 H2S(g) D 2 H2 (g) + S2 (g), Kp = 1.5 × 10−5 at 800.0°C. If the initial partial pressures of H2 and S2 in a cl
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Answer: The approximate equilibrium partial pressure of H_2S is 3.92  atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

      2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)

K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}

1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}

On reversing the reaction:

     2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)

initial pressure  4.00atm    2.00 atm       0

eqm          (4.00-2x)atm      (2.00-x) atm      2x atm

K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}

K_p'=\frac{1}{K_p}=0.67\times 10^5

2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)

0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}

x=1.96

[H_2S]=2x=2\times 1.96=3.92 atm

Thus approximate equilibrium partial pressure of H_2S is 3.92 atm

3 0
3 years ago
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