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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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Write the balanced equation for the ionization of the weak base pyridine, c5h5n , in water, h2o. Phases are optional.
Lelu [443]

The balanced equation for the ionization of the weak base pyridine,C5H5N in water, H2O

C_5H_5N ( aq.) + H2O ( l) ---------> C5H5NH+ (aq.) + OH- (aq.)

<h3>What is the balanced equation for the ionization?</h3>

Generally, Pyridine is characterized by a ring structure, in this characteristic ring structure N is sp2 hybridized, hence creating a lone pair present on N so s - character is more, as well as lone pair, is present.

Therefore, Considering The following functions of the equation:weak base pyridine,C5H5N in water, H2O

We write the balanced equation for the ionization as

C_5H_5N ( aq.) + H2O ( l) ---------> C5H5NH+ (aq.) + OH- (aq.)

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6 0
2 years ago
Discuss elements and compounds by completing the following paragraph
pychu [463]

Elements are substances that are made up of the same atoms which are capable of taking part in a chemical reaction.

There are different types of elements which are represented by symbols gotten from the first letter or the first and any other letter in the name of the element.

Examples of elements include:

  • Hydrogen (H)
  • Carbon (C)
  • Nitrogen (N)
  • Sodium (Na)

When two or more of these elements combine together through a chemical bond, it leads to the formation of compounds.

Example of a compound includes:

  • NaCl: The element sodium combine, through electrochemical bonding, with another element chlorine to form the compound sodium chloride.

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3 0
2 years ago
Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an
ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

TiCl_{4(l)} ⇒ TiCl_{4(g)}

ΔH°_{f} (TiCl_{4(l)}) = -804.2 kJ/mol

ΔH°_{f} (TiCl_{4(g)}) = -763.2 kJ/mol

Therefore,

ΔH°_{f} = ΔH°_{f} (TiCl_{4(g)}) - ΔH°_{f} (TiCl_{4(l)}) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(TiCl_{4(l)}) = 221.9 J/(mol*K)

s°(TiCl_{4(g)}) = 354.9 J/(mol*K)

Therefore,

s° = s° (TiCl_{4(g)}) - s°(TiCl_{4(l)}) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH°_{f} /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

5 0
3 years ago
Comprised of protons, neutrons, and electrons.
dsp73

Answer:

The atom

Explanation:

6 0
3 years ago
Calculate the activity coefficients for the following conditions:
uysha [10]

<u>Answer:</u>

<u>For a:</u> The activity coefficient of copper ions is 0.676

<u>For b:</u> The activity coefficient of potassium ions is 0.851

<u>For c:</u> The activity coefficient of potassium ions is 0.794

<u>Explanation:</u>

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}       ........(1)

where,

\gamma_i = activity coefficient of ion

Z_i = charge of the ion

\mu = ionic strength of solution

\alpha _i = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)        ......(2)

where,

C_i = concentration of i-th ions

Z_i = charge of i-th ions

  • <u>For a:</u>

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M

Now, calculating the activity coefficient of Cu^{2+} ion in the solution by using equation 1:

Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{  (known)}\\\mu=0.01M

Putting values in equation 1, we get:

-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676

Hence, the activity coefficient of copper ions is 0.676

  • <u>For b:</u>

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M

Now, calculating the activity coefficient of K^{+} ion in the solution by using equation 1:

Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{  (known)}\\\mu=0.025M

Putting values in equation 1, we get:

-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851

Hence, the activity coefficient of potassium ions is 0.851

  • <u>For c:</u>

We are given:

0.02 M K_2SO_4 solution:

Calculating the ionic strength by using equation 2:

C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M

Now, calculating the activity coefficient of K^{+} ion in the solution by using equation 1:

Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{  (known)}\\\mu=0.06M

Putting values in equation 1, we get:

-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794

Hence, the activity coefficient of potassium ions is 0.794

6 0
3 years ago
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