Answer:
-255.4 kJ
Explanation:
The free energy of a reversible reaction can be calculated by:
ΔG = (ΔG° + RTlnQ)*n
Where R is the gas constant (8.314x10⁻³ kJ/mol.K), T is the temperature in K, n is the number of moles of the products (n =1), and Q is the reaction quotient, which is calculated based on the multiplication of partial pressures by the partial pressure of the products elevated by their coefficient divide by the multiplication of the partial pressure of the reactants elevated by their coefficients.
C₂H₂(g) + 2H₂(g) ⇄ C₂H₆(g)
Q = pC₂H₆/[pC₂H₂ * (pH₂)²]
Q = 0.261/[8.58*(3.06)²]
Q = 3.2487x10⁻³
ΔG = -241.2 + 8.314x10⁻³x298*ln(3.2487x10⁻³)
ΔG = -255.4 kJ
where are the element you haven't showed it so that we can answer it
<u>Answer:</u> The final volume of the gas comes out to be 4 L.
<u>Explanation:</u>
To calculate the volume with changing pressure, we use the equation given by Boyle's law.
This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
Mathematically,
(At constant temperature and number of moles)
The equation given by this law is:
where,
are initial pressure and volume.
are final pressure and volume.
We are given:
Putting values in above equation, we get:
Hence, the final volume of the gas will be 4 L.
Answer:
Iodide> Bromide > chloride > flouride
Explanation:
During a nucleophilic substitution reaction, a nucleophilie replaces another in a molecule.
This process may occur via an ionic mechanism (SN1) or via a concerted mechanism (SN2).
In either case, the ease of departure of the leaving group is determined by the nature of the C-X bond. The stronger the C-X bond, the worse the leaving group will be in nucleophilic substitution. The order of strength of C-X bond is F>Cl>Br>I.
Hence, iodine displays the weakest C-X bond strength and it is thus, a very good leaving group in nucleophillic substitution while fluorine displays a very high C-X bond strength hence it is a bad leaving group in nucleophilic substitution.
Therefore, the ease of the use of halide ions as leaving groups follows the trend; Iodide> Bromide > chloride > flouride
