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ludmilkaskok [199]
3 years ago
9

What are the partial pressures of NH3 and H2S at equilibrium, that is, what are the values of PNH3 and PH2S, respectively?Enter

the partial pressure of ammonia followed by the partial pressure of hydrogen sulfide numerically in atmospheres separated by a comma.
Chemistry
1 answer:
mylen [45]3 years ago
6 0

Answer:

Ammonium bisulfide, NH4HS , forms ammonia, NH3 , and hydrogen sulfide, H2S , through the reaction NH4HS(s)⇄NH3(g)+H2S(g) This reaction has a Kp value of 0.120 at 25°C .

An empty 5.00-L flask is charged with 0.300 g of pure H2S(g) , at 25°C

Partial pressure of NH3 = 0.325 atm

Partial pressure of H2S  = 0.368 atm

Explanation:

Temperature = 25 °C = 298 K

Volume = 5 L

Mass of H2S = 0.30 g

Moles of H2S = 0.30 / 34.08

= 0.0088

Using PV = nRT

Initial pressure of H2S = 0.0088 * 0.0821 * 298 / 5

= 0.043 atm

Kp = 0.120

(a). NH4HS(s) ⇄ NH3(g) + H2S(g)

Initial - 0.0 0.043

Change - + x + x

Equilibrium - x 0.043 + x

Kp = P(NH3) * P(H2S)

0.120 = x (0.043 + x)

x = 0.325 atm

Hence, at equilibrium:

Partial pressure of NH3 = 0.325 atm

Partial pressure of H2S = 0.043 + 0.325 = 0.368 atm

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1. How many molecules of S2 gas are in 756.2 L?
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Answer: There are 2.032 \times 10^{25} molecules S_{2} gas are in 756.2 L.

Explanation:

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According to mole concept, 1 mole of every substance contains 6.022 \times 10^{23} molecules.

Therefore, molecules of S present in 33.76 moles are calculated as follows.

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5 0
3 years ago
The concentrations of the products at equilibrium are [pcl3] = 0.180 m and [cl2] = 0.250 m. what is the concentration of the rea
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We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula: 
Kc   = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 =  0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07

4 0
3 years ago
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