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allsm [11]
3 years ago
13

Nitrogen monoxide and water react to form ammonia and oxygen, like this: (g)(g)(g)(g) Write the pressure equilibrium constant ex

pression for this reaction.
Chemistry
1 answer:
xxTIMURxx [149]3 years ago
6 0

Answer:

Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}

Explanation:

First, we have to write the balanced chemical equation for the reaction. Nitrogen monoxide (NO) reacts with water (H₂O) to give ammonia (NH₃) and oxygen (O₂), according to the following:

NO(g) + H₂O(g) → NH₃(g) + O₂(g)

To balance the equation, we add the stoichiometric coefficients (4 for NH₃ and NO to balance N atoms, then 6 for H₂O to balance H atoms and then 5 for O₂ to balance O atoms):

4 NO(g) + 6 H₂O(g) → 4 NH₃(g) + 5 O₂(g)

All reactants and products are in the gaseous phase, so the equilibrium constant is expressed in terms of partial pressures (P) and is denoted as Kp. The Kp is expressed as the product of the reaction products (NH₃ and O₃) raised by their stoichiometric coefficients (4 and 5, respectively) divided into the product of the reaction reagents (NO and H₂O) raised by their stoichiometric coefficients (4 and 6, respectively). So, the pressure equilibrium constant expression is written as follows:

Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}

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4 years ago
Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2
Aleksandr-060686 [28]

Answer : The H_3O^+ ion concentration is, 1.12\times 10^{-3}M and the pH of a buffer is, 2.95

Explanation : Given,

K_a=7.1\times 10^{-4}

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Concentration of KNO_2 (conjugate base or salt)= 0.89 M

First we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (7.1\times 10^{-4})

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Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}

Now put all the given values in this expression, we get:

pH=3.15+\log (\frac{0.89}{0.26})

pH=2.95

The pH of a buffer is, 2.95

Now we have to calculate the H_3O^+ ion concentration.

pH=-\log [H_3O^+]

2.95=-\log [H_3O^+]

[H_3O^+]=1.12\times 10^{-3}M

The H_3O^+ ion concentration is, 1.12\times 10^{-3}M

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the percentage of heaviest Sn is 5.80%

the given mass of Sn is 82g

The total  moles of Sn will be = mass / atomic mass = 82/118.71=0.691

Total atoms of Sn in 82g = 6.023X10^{23}X0.691=4.16X10^{23}

the percentage of heaviest Sn is 5.80%

So the total atoms of Sn^{124} = 5.80% X 4.16X10^{23}

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Answer:

.

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6 0
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