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grin007 [14]
3 years ago
5

When a chemical reaction reaches chemical equilibrium which occurs faster, the formation of product from reactants or reactants

from product?
Chemistry
1 answer:
nekit [7.7K]3 years ago
4 0

Answer: when a system reaches equilibrium, the rate of formation of product equals the rate of formation of reactant. At this state, we say the reaction is in dynamic equilibrium. Therefore the rate of formation of product and the rate of formation of reactant are equal

Explanation:

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If all other variables remain unchanged, what happens to the output force when the area of the input piston is doubled?
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Answer:

1250N

Explanation:

This question is based on pascal's Law.

So By Pascal's Law

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therefore after putting the values we get,

= (25x 1500)/30

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What is true about energy in an ordinary chemical reaction? Energy is
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The answer is statement #3.
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CO(g)+2H2(g)⇌CH3OH(g)CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial concentrations o
Katarina [22]

Answer:

9.4

Explanation:

The equation for the reaction can be represented as:

CO_{(g)}    +      2H_2O_{(g)}   ⇄  CH_3OH_{(g)}

The ICE table can be represented as:

                                  CO_{(g)}    +      2H_2_{(g)}   ⇄  CH_3OH_{(g)}

Initial                          0.27             0.49              0.0

Change                      -x                  -2x                 x

Equilibrium               0.27 - x         0.49 -2x          x

We can now say that the concentration of  CH_3OH_{(g)} at equilibrium is x;

Let's not forget that at equilibrium  CH_3OH_{(g)} = 0.11 M

So:

x =  [CH_3OH_{(g)}] = 0.11 M

[CO_{(g)}] = 0.27 - x

[CO_{(g)}] = 0.27 - 0.11

[CO_{(g)}] = 0.16 M

[2H_2_{(g)}] = (0.49 - 2x)

[2H_2_{(g)}] = (0.49 - 2(0.11))

[2H_2_{(g)}] = 0.49 - 0.22

[2H_2_{(g)}] = 0.27 M

K_C = \frac{[CH_3OH]}{[CO][H_2]^2}

K_C = \frac{(0.11)}{(0.16)[(0.27)^2}

K_C = 9.4307

K_C = 9.4

∴ The equilibrium constant at that temperature = 9.4

8 0
3 years ago
In three different experiments, the following results were obtained for the reaction
faust18 [17]
From the given observations,
You can see that as the concentration is doubled, half-life is halved.

That is,half-life is inversely proportional to concentration

As t( half-life) ~ 1/a^(n-1)

For this case n = 2,second order reaction.

R = k X a^n

Using the above formula you will get the rate and rate constant.
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