1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
erik [133]
3 years ago
6

The equilibrium constant values for the reaction of Ni+2 in aqueous solution with an amine to form the corresponding ammonia, en

, and dien complexes increase in the order dien>en>ammonia. Why is this the case?
Chemistry
1 answer:
Alex_Xolod [135]3 years ago
4 0

Answer:

Relative crystal field splitting of the ligands

Explanation:

The formation constant of a complex is the equilibrium constant that shows the stability of the complex. This equilibrium constant is found to depend on the identity of the central metal atom/ion and the identity of the ligands.

Some ligands are called strong field ligands while others are known as weak field ligands. Other ligands lie somewhere in between the two extremes in the spectrochemical series. The spectrochemical series arranges ligands in order of their relative magnitude of crystal field splitting. The greater the magnitude of crystal field splitting, the greater the formation constant or equilibrium constant of the complex.

Strong field ligands are able to participate in ligand to metal charge transfer or metal to ligand charge transfer as the case may be. Hence strong field ligands possess pi-orbitals which interact with metal pi orbitals.

The order of decreasing formation constant in the question reflects the relative positions of the ligands in the spectrochemical series. Ammonia is lower than ethylene diammine in the series which is also lower than a diene. A diene has pi orbitals capable of interaction with metal pi orbitals hence the highest crystal field splitting and equilibrium constant.

You might be interested in
i am begging anyone to help me with this! (all tutors i've asked said they can't solve it but i need someone to help me out) - i
9966 [12]

First, we need to calculate how much energy we will get from this combustion.

Assuming the combustion is complete, we have the octane reacting with O₂ to form only water and CO₂, so:

C_8H_{18}+O_2\to CO_2+H_2O

We need to balance the reaction. Carbon only appear on two parts, so, we can start by it:

C_8H_{18}+O_2\to8CO_2+H_2O

Now, we balance the hydrogen:

C_8H_{18}+O_2\to8CO_2+9H_2O

And in the end, the oxygen:

C_8H_{18}+\frac{25}{2}O_2\to8CO_2+9H_2O

We can multiply all coefficients by 2 to get integer ones:

2C_8H_{18}+25O_2\to16CO_2+18H_2O

Now, we need to use the enthalpies of formation to get the enthalpy of reaction of this reaction.

The enthalpy of reaction can be calculated by adding the enthalpies of formation of the products multiplied by their stoichiometric coefficients and substracting the sum of enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.

For the reactants, we have (the enthalpy of formation of pure compounds is zero, which is the case for O₂):

\begin{gathered} \Delta H\mleft\lbrace reactants\mright\rbrace=2\cdot\Delta H\mleft\lbrace C_8H_{18}\mright\rbrace+25\cdot\Delta H\mleft\lbrace O_2\mright\rbrace \\ \Delta H\lbrace reactants\rbrace=2\cdot(-250.1kJ)+25\cdot0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ+0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ \end{gathered}

For the products, we have:

\begin{gathered} \Delta H_{}\mleft\lbrace product\mright\rbrace=16\cdot\Delta H\lbrace CO_2\rbrace+18\cdot\Delta H\lbrace H_2O\rbrace \\ \Delta H_{}\lbrace product\rbrace=16\cdot(-393.5kJ)+18\cdot(-285.5kJ) \\ \Delta H_{}\lbrace product\rbrace=-6296kJ-5139kJ \\ \Delta H_{}\lbrace product\rbrace=-11435kJ \end{gathered}

Now, we substract the rectants from the produtcs:

\begin{gathered} \Delta H_r=\Delta H_{}\lbrace product\rbrace-\Delta H\lbrace reactants\rbrace \\ \Delta H_r=-11435kJ-(-500.2kJ) \\ \Delta H_r=-10934.8kJ \end{gathered}

Now, this enthalpy of reaction is for 2 moles of C₈H₁₈, so for 1 mol of C₈H₁₈ we have half this value:

\Delta H_c=\frac{1}{2}\Delta H_r=\frac{1}{2}\cdot(-10934.8kJ)=-5467.4kJ

Now, we have 100 g of C₈H₁₈, and its molar weight is approximately 114.22852 g/mol, so the number of moles in 100 g of C₈H₁₈ is:

\begin{gathered} M_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{n_{C_8H_{18}}} \\ n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{100g}{114.22852g/mol}\approx0.875438mol \end{gathered}

Since we have approximately 0.875438 mol, and 1 mol releases -5467.4kJ when combusted, we have:

Q=-5467.4kJ/mol\cdot0.875438mol\approx-4786.37kJ

Now, for the other part, we need to calculate how much heat it is necessary to melt a mass, <em>m</em>.

First, we have to heat the ice to 0 °C, so:

\begin{gathered} Q_1=m\cdot2.010J/g.\degree C\cdot(0-(-10))\degree C \\ Q_1=m\cdot2.010J/g\cdot10 \\ Q_1=m\cdot20.10J/g \end{gathered}

Then, we need to melt all this mass, so we use the latent heat now:

Q_2=n\cdot6.03kJ/mol

Converting mass to number of moles of water we have:

\begin{gathered} M=\frac{m}{n} \\ n=\frac{m}{M}=\frac{m}{18.01528g/mol} \end{gathered}

So:

Q_2=\frac{m}{18.01528g/mol}_{}\cdot6.03kJ/mol\approx m\cdot0.334716kJ/g

Adding them, we have a total heat of:

\begin{gathered} Q_T=m\cdot20.10J/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.02010kJ/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.354816kJ/g \end{gathered}

Since we have a heat of 4786.37 kJ form the combustion, we input that to get the mass (the negative sign is removed because it only means that the heat is released from the reaction, but now it is absorbed by the ice):

\begin{gathered} 4786.37kJ=m\cdot0.354816kJ/g \\ m=\frac{4786.37kJ}{0.354816kJ/g}\approx13489g\approx13.5\operatorname{kg} \end{gathered}

Since we have a total of 20kg of ice, we can clculate the percent using it:

P=\frac{13.5\operatorname{kg}}{20\operatorname{kg}}=0.675=67.5\%

5 0
1 year ago
A person weighed 15 pennies on a balance and recorded the following masses:
Artist 52 [7]

Answer:

A)  yes

B) The average mass of 12 pennies should be expressed as

( 3.131 + 3.129 + -------- X12 ) g /  12

where X12 = is the mass of the 12th penny

Explanation:

A) I think the the Bureau of Mint changed the way it made pennies because from the experiment and observation carried out on 15 different pennies it can be seen that they had different weights and this difference is associated with the period/time of production of the different pennies,

B ) The average mass of any object should be expressed in the si unit of mass and not having ($) sign attached to the value instead it should be expressed in grams or kilograms (s.i unit of mass )

The average mass of 12 pennies should be expressed as

( 3.131 + 3.129 + -------- X12 ) g /  12

where X12 = is the mass of the 12th penny

4 0
3 years ago
If the temperature of a gas increased and the pressure remains the same what happens to the volume of gas?​
podryga [215]

Answer:

the answer is B.

Explanation:

The pressure increases... plus i had this same question

pls give me brainliest

3 0
3 years ago
Which change does an increase in heat energy cause?
Bumek [7]

Answer:

C more collision between molecules

Explanation:

increase in heat causes increase in kinetic energy of the particles

5 0
2 years ago
What is the main benefit of this model?
igor_vitrenko [27]

Answer:

letter D.

Explanation:

it shows a pattern for determining the side of lengths of a triangle yeah the triangle had a model and a length

4 0
3 years ago
Other questions:
  • Questions 18-20 refer to the reaction 2NH3 (g) ↔ N2 (g) + 3H2 (g), and ΔH = 92.2 kJ.
    7·1 answer
  • Suppose a platinum atom in the oxidation state formed a complex with two chloride anions and two ammonia molecules. Write the ch
    13·1 answer
  • Which individual elements make up the
    9·1 answer
  • How many atoms of lead are contained in 0.532mol?​
    12·1 answer
  • How many molecules are in 15 grams of Oxygen gas?
    13·1 answer
  • 1 lb of CO2 occupies 0.6 ft^3 at a pressure of 200 psi. Determine the temperature of the system.
    12·1 answer
  • What is the maximum amount of NaCl that can dissolve in 300 g of water? (the solubility of NaCl is 36.2 g/100 g H2O at 20 ⁰C.) p
    8·1 answer
  • Use Hess law and the following equations to calculate H for the reactio below
    7·1 answer
  • Please help me it is due in 10 min <br> there is 8 questions
    9·1 answer
  • The electrophilic bromination or chlorination of benzene requires ______ along with the halogen.
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!