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balu736 [363]
3 years ago
14

Which of the following statements is FALSE?

Chemistry
2 answers:
pshichka [43]3 years ago
3 0

Answer:

D

Explanation:

D. K >> 1 implies that the reaction is very fast at producing products.

oee [108]3 years ago
3 0

Answer:

(D) K >> 1 implies that the reaction is very fast at producing products.

Explanation:

For a reaction:

nA + xB ⇄ bC + aD

The thermodynamic equilibrium constant (k) is:

k = \frac{[C]^b[D]^a]}{[A]^n[B]^x} <em>(1)</em>

(A) When K >> 1, the forward reaction is favored and essentially goes to completion.  <em>TRUE. </em>By (1), a k>> 1 means that you have more products than reactants

(B) When K << 1, the reverse reaction is favored and the forward reaction does not proceed to a great extent.  <em>TRUE. </em>By (1), a k<< 1 means that you have more reactants than products.

(C) When K ≈ 1, neither the forward or reverse reaction is strongly favored, and about the same amount of reactants and products exist at equilibrium.  <em>TRUE. </em>By (1), a k ≈ 1 means that you have the same concentration of products andreactants

(D) K >> 1 implies that the reaction is very fast at producing products.  <em>FALSE. </em>k is a thermodynamic parameter. To evaluate the rate of a reaction you need to study the kinetics of this one.

I hope it helps!

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Hi guys, my question is:
guapka [62]

Answer:

See Explanation Below

Explanation:

A) The rate law can only be on the reactant side and you can only determine it after you get the net ionic equation because of spectators cancelling out. So in this case the rate law is k=[CH3Br]^1 [OH-]^1. The powers are there because the rxn is first order.  

B) Since the rxn is first order anything you do to it will be the exact same "counter rxn" per say so since you are decreasing the OH- by 5 the rate will decease by 5

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Explanation:

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3 years ago
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Klio2033 [76]

Rate equation for first order reaction is as follows:

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}

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t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days

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(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

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(c)  Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

5 0
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7 0
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