Do you see patterns in the distribution of their electron?

During transpeptidation, nucleophilic attack occurs between the α-amine group in the A site and the esterified carbon in the P site

<h3>What is transpeptidation ?</h3>

A chemical process in which an amino acid residue or peptide residue is transferred from one amino compound to another. This process occurs when a protease reversibly converts one peptide to another.

a process where one or more amino acids are transferred from one peptide chain to another, as in the case of a transpeptidase activity, or where a peptide chain itself is produced, as in the case of the production of a bacterial cell wall.
Peptidoglycan glyco-syltransferases create the glycan strands (PGTs). The most popular class of antibiotics' fatal targets are enzymes termed transpeptidases (TPs), which produce the peptide crosslinks: using beta-lactams

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1 year ago

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

Lerok [7]

Answer:

Approximately $\rm -89 \; kJ\cdot mol^{-1}$ .

Assumption: the density of the solution is equal to the density of pure water.

Explanation:

The enthalpy of neutralization is defined as the enthalpy change for each moles of water produced. (Clark, Physical & Theoretical Chemistry, Chemistry Libretexts.)

Each mole of $\rm NaOH$ formula units will neutralize one mole of $\rm HCl$ to produce one mole of water. $\rm HCl$ and $\rm NaOH$ are available at equal volume and concentration. In other words, there's an equal number of both reactants. All $\rm HCl$ and $\rm NaOH$ will react to form water.

$V(\mathrm{HCl}) = \rm 137\; cm^{3} = 0.137\;dm^{3}$ .

$V(\mathrm{NaOH}) = \rm 137\; cm^{3} = 0.137\;dm^{3}$ .

$n = c\cdot V = \rm 0.137\;dm^{3} \times 2.6\;mol\cdot dm^{-3} = 0.3652\; mol$ .

In other words, there are $\rm 0.3652\; mol$ of $\rm HCl$ and $\rm NaOH$ each. The two will react to produce $\rm 0.3652\; mol$ of water.

How much heat is released?

Assume that the volume of the liquid is equal to the volume of the $\rm HCl$ solution plus the volume of the $\rm NaOH$ solution. That's $\rm 0.274\;dm^{3}$ . Assume that the density of the solution is equal to that of water under room temperature. $\rho(\text{water}) = \rm 1.000\; kg\cdot dm^{-3}$ . The mass of the liquid will be $m = \rho \cdot V = \rm 0.274\; dm^{3} \times 1.000\; kg\cdot dm^{-3} = 0.274\; kg = 274\;g$ .

Change in temperature:

$\Delta T = \rm 325.8 - 298 = 27.8\; K$ .

Heat that the solution absorbed:

$Q = c\cdot m \cdot \Delta T = \rm 4.18\;J\cdot K^{-1}\cdot g^{-1} \times 274\; g\times 27.8\;K = 36410.216\; J = 36.410216\; kJ$ .

That will also be the amount of heat released from the reaction if there's no energy loss.

$\displaystyle \Delta H(\text{Neutralization}) = \frac{-Q}{n(\text{water produced})} = \rm \frac{36.410216\; kJ}{0.3652\; mol} \approx 89\; kJ\cdot mol^{-1}$ .

6 0

3 years ago