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goldenfox [79]
3 years ago
13

Discuss: Reaction-Rate Factors

Chemistry
1 answer:
SpyIntel [72]3 years ago
3 0
1) The time depends on what the lab wanted you to do.  It will tell you in the procedure when you are supposed to considered a reaction to be complete and you just measure the time for that to happen.

2) Most text books say that increasing the concentration of one or more reactants will increase the rate of the reaction.  To prove this with your data you need to show that when you increased the concentration of one of the reactants, the reaction rate did increase.  The results of this experiment are not enough to make a general statement since the experiment was not on a large enough scale to diffidently prove anything.  (you could have been testing the one exception or had a error in one of your trials)
I hope this helps.  Let me know in the comments if anything is unclear.

(The concentration of one or more of the reactants will increase the rate of the reaction.  This is explained through the fact that all reactions require collisions that have certain orientations and a minimum energy level.  By increasing the concentration of one or more reactants, you   increase the number of collisions which increases the rate since requires collisions in order to occur.)  <span />
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<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

CaCl₂ + 2AgNO₃ → 2AgCl + Ca(NO₃)₂

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\tt \dfrac{2}{169.9}=0.012

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Black_prince [1.1K]

Answer:

The correct option is: A) 2H₂(g) + O₂(g) → 2H₂O(g)

Explanation:

According to the Le Chatelier's principle, change in the volume of the reaction system causes equilibrium to shift in the direction that reduces the effect of the volume change.

When the <u>volume decreases then the pressure of the reaction vessel increases, then the equilibrium shifts towards the reaction side that produces less number of moles of gas.</u>

<u />

A) 2H₂(g) + O₂(g) → 2H₂O(g)

The number of moles of reactant is 3 and number of moles of product is 2.

<u>Therefore, when volume decreases, the equilibrium shifts towards the product side, thereby </u><em><u>favoring the formation of products.</u></em>

B) NO₂(g) + CO(g) → NO(g) + CO₂(g)

The number of moles of reactant and product both is 2.

<u>Therefore, when the volume decreases, the equilibrium does not shift in any direction.</u>

C) H₂(g) + I₂(g) → 2HI(g)

The number of moles of reactant and product both is 2.

<u>Therefore, when the volume decreases, the equilibrium does not shift in any direction.</u>

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The number of moles of reactant is 2 and number of moles of product is 3.

<u>Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby </u><em><u>favoring the formation of reactants.</u></em>

E) MgCO₃(s) → MgO(s) + CO₂(g)

The number of moles of reactant is 1 and number of moles of product is 2.

<u>Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby </u><em><u>favoring the formation of reactants.</u></em>

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