Consider the following reaction:

Chemistry

1 answer:

rusak2 [61]4 years ago

8 0

Answer:

1. a) 0,18

2. e) The reaction proceeds to the left, forming more BrCl(g).

Explanation:

1. In a gas reaction as:

2SO₂(g) +O₂(g) ⇄ 2SO₃(g)

it is possible to convert kp to kc using:

kp = kc (RT)^Δn

Where kp is gas equilibrium constant, kc is equilibrium constant (13), R is gas constant (0,082atmL/molK), T is temperature (900K), and Δn is number of moles of gas products - number of moles of gas reactants (That is 2 - (2+1) = -1). Replacing:

kp = 13×(0,082atmL/molK×900K)^-1

kp = 0,18

2. Based on Le Chatelier's principle, the change in temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system. For the reaction:

2 BrCl(g) ⇄ Br₂(g) + Cl₂(g).

The addition of 0,050M of each compound cause the reaction proceeds to the left, forming more BrCl(g) because based on the reaction, you need two moles of BrCl per mole of Br₂(g) and Cl₂(g) to keep the system in the same. But you are adding the same proportion of moles of each compound.

I hope it helps!

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A polar bond is formed with atoms having different electronegativities. The bonding electrons are attracted more towards the atom with greater electronegativity resulting in unequal sharing of electrons. Therefore the molecule develop partial charges and becomes polar. Polar molecules have dipole moment that is the partial charge on molecules due to differences in electronegativity between atoms.

A non-polar bond is formed with atoms having the same electronegativity, hence the bonded pair of electron is shared equally between atoms. Non-polar molecules have no moment.
Note that: symmetrical molecules having polar bonds are non-polar because the dipoles of the bond exert equal and opposite effect. Hence the dipoles cancel the charges.

Example: HCl
In HCl, Cl is more electronegative therefore Cl atom pulls the electron pair of the covalent bond towards itself and develops a partial negative charge. Consequently H develops a partial positive charge. This therfore leads to the formation of a dipole.