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Svetradugi [14.3K]
2 years ago
12

What is the numerical value for Q for this reaction? Remember to always express your answer to correct number of significant fig

ures.

Chemistry
1 answer:
ioda2 years ago
3 0

The reaction is not at equilibrium because Q >P hence the reaction moves towards the left hand side.

<h3>What is the equilibrium constant?</h3>

The equilibrium constant is a figure that tells us the extent to which the reactants have been converted to products in a reaction.

Now we have the reaction; PCl3(g) + Cl2(g) ⇄ PCl5(g) hence we can write;

Q = [PCl5]/[PCl3] [Cl2]

Q = 0.39/0.21 * 0.41

Q = 4.5

The reaction is not at equilibrium because Q >P hence the reaction moves towards the left hand side.

Learn more about equilibrium constant:brainly.com/question/10038290

#SPJ1

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andreev551 [17]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

rate=k[H_2]^2[I_2]^2

At a certain concentration of H_2 and I_2, the initial rate of reaction is 4.0 × 10⁴ M/s. What would the initial rate of the reaction be if the concentration of

Answer : The initial rate of the reaction will be, 1.0\times 10^4M/s  

Explanation :

Rate law expression for the reaction:

rate=k[H_2]^2[I_2]^2

As we are given that:

Initial rate = 4.0 × 10⁴ M/s

Expression for rate law for first observation:

4.0\times 10^4=k[H_2]^2[I_2]^2 ....(1)

Expression for rate law for second observation:

R=k(\frac{[H_2]}{2})^2[I_2]^2 ....(2)

Dividing 2 by 1, we get:

\frac{R}{4.0\times 10^4}=\frac{k(\frac{[H_2]}{2})^2[I_2]^2}{k[H_2]^2[I_2]^2}

\frac{R}{4.0\times 10^4}=\frac{1}{4}

R=1.0\times 10^4M/s

Therefore, the initial rate of the reaction will be, 1.0\times 10^4M/s

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