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Kamila [148]
3 years ago
7

The elementary reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pres

sures of H2OH2O , H2H2 , and O2O2 reach 0.070 atm,0.070 atm, 0.0035 atm,0.0035 atm, and 0.0025 atm,0.0025 atm, respectively at equilibrium. What is the value of the equilibrium constant at this temperature?
Chemistry
1 answer:
Dima020 [189]3 years ago
5 0

Answer:

6.25\times 10^{-6} is the value of the equilibrium constant at this temperature.

Explanation:

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures  of reactants each raised to the power equal to their stoichiometric ratios. It is expressed as K_{p}

2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)

Partial pressures at equilibrium:

p^o_{H_2O}=0.070 atm

p^o_{H_2}=0.0035 atm

p^o_{O_2}=0.0025 atm

The equilibrium constant in terms of pressures is given as:

K_p=\frac{(p^o_{H_2})^2\times (p^o_{O_2})}{(p^o_{H_2O})62}

K_p=\frac{(0.0035 atm)^2\times 0.0025 atm}{(0.070 atm)^2}=6.25\times 10^{-6}

6.25\times 10^{-6} is the value of the equilibrium constant at this temperature.

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If 4.12 l of a 0.850 m-h3po4 solution is be diluted to a volume of 10.00 l, what is the concentration of the resulting solution?
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The process in which the concentration of the solution is lessened  by the addition of water is said to be dilution and equation of dilution relates the initial concentration and volume of stock solution with the final concentration and volume of the solution.

Formula is given by:

C_{1}V_{1}=C_{2}V_{2}   (1)

where,

C_{1} is the initial concentration

V_{1} is the initial volume

C_{2} is the final concentration

V_{2} is the final volume

Now,

C_{1} = 0.850 M

V_{1} = 4.12 L

C_{2}  =?

V_{2} = 10.00 L

Substitute the give values in formula (1),

0.850 M\times 4.12L=C_{2}\times 10.00 L

C_{2} =\frac{0.850 M\times 4.12L}{10.00 L}

= 0.3502 M

Thus, the final concentration of theH_{3}PO_{4} solution = 0.3502 M












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3 years ago
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