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kodGreya [7K]
3 years ago
15

Explain what changes and what stays the same 1.00 L of a solution of NaCl is diluted to 1.80 L

Chemistry
1 answer:
Reika [66]3 years ago
7 0

Explanation:

A process in which there occurs a decrease in concentration of solute particles upon addition of more amount of solvent is known as dilution.

Also, we know that molarity is the number of moles present in a liter of solution.

Mathematically,      Molarity = \frac{\text{no. of moles}}{\text{Volume in liter}}

So, more is the volume of solution lesser will be the number of moles present in it. This means that when we increase the amount of solvent then there will be decrease in concentration of solute particles.

Thus, when 1.00 L of a solution of NaCl is diluted to 1.80 L then its molarity will decrease due to decrease in number of moles of NaCl.

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FILL IN THE BLANKWORD BANK (can use more than once): less, increases, decreases, greaterNO2Cl(g)+NO(g)⇌NOCl(g)+NO2(g)1.Disturbin
Jlenok [28]

Explanation:

If we have the following reaction at equilibrium:

<em>                                           aA + bB ⇄ cC + dD</em>

where a, b, c, and d are the stoichiometric coefficients for the reacting species A, B,  C, and D. For the reaction at a particular temperature:

                       Kc=([C]^c *[D]^d)/([A]^a *[B]^b)

where Kc is the equilibrium constant, which holds that <em>for a reversible reaction at equilibrium  and a constant temperature, a certain ratio of reactant and product concentrations has  a constant value, Kc</em> (the equilibrium constant). Note that although the concentrations  may vary, as long as the reaction in in equilibrium and temperatura don't change, the value of <u>K remains constant.</u>

For reactions that have not reached equilibrium, we obtain the reaction quotient (Qc), instead of the equilibrium  constant <u>by substituting the initial concentrations into the equilibrium constant expression.</u>

                        Qc=([Co]^c *[Do]^d)/([Ao]^a *[Bo]^b)

To determine the direction in wich the net reaction will proceed to reach equilibrium, que compare the values of Qc and Kc.

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  • Qc = Kc: The initial concentrations are equilibrium concentrations. The system in at equilibrium.
  • Qc > Kc: To reach equilibrium, products must be converted to reactants (←)

Solution:

We have the following reaction:

                             NO2Cl(g)+NO(g)⇌NOCl(g)+NO2(g)

So:

Kc=([NOCl]^1*[NO2]^1)/([NO2Cl]^1 *[NO]^1)

     =([NOCl][NO2])/([NO2Cl][NO])  

1. In the equation above, [NO2Cl] it's in the denominator, so if we increase it's numericall value by adding NO2Cl <u>decreases Qc  to a value less than Kc.</u>

<em>(From the chemical point of view, if we disturb the equilibrium adding NO2Cl (a reactant), to reach equilibrium again the system proceeds from left to right (→) consuming this reactant.)</em>

2. To reach a new state of equilibrium (<em>where Qc = Kc</em>), Qc therefore  increases wich means that the denominator of the expression for Qc  decreases <em>(in order to increase the denominator as mention above).</em>

3. To accomplish this, the concentration of reageants decreases <em>(reagents are being consumed), </em>and the concentration of prodcuts increases <em>(products are being formed).</em>

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