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mina [271]
3 years ago
9

What happens when the temperature or pressure of a reaction system at equilibrium is altered? The equilibrium state is unable to

reestablish. The position of equilibrium is maintained. The system will try and offset the change. The system will soon reach a state of chaos.
Chemistry
1 answer:
alina1380 [7]3 years ago
3 0

Answer: The system will try and offset the change.

Explanation: Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in a direction to minimize the effect.

Thus if temperature is increased, the reaction will shift in a direction where temperature is decreasing and vice versa. Similarly if pressure is increased, the reaction will shift in a direction where pressure is decreasing and vice versa.

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Based on the activity series of metal, which reaction with water will not happen?
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How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56
neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of [A_0] is decomposed. So,

\frac {[A_t]}{[A_0]} = 0.0156

Thus,

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.0156=e^{-k\times t}

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}

t = 6\times t_{1/2}

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

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The difference in elevation between the highest and lowest contour lines on a topographical map is called:
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