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zubka84 [21]
3 years ago
15

Chemical weathering occurs faster in warm, tropical climates, although it happens in cold climates as well. Describe some condit

ions in Antarctica that could lead to chemical weathering.
Chemistry
2 answers:
vredina [299]3 years ago
7 0

Answer:

Oxidation, acidification and hydrolisis.

Explanation:

Hello,

It is widely known that chemical weathering occurs when rocks undergo chemical reactions to yield different minerals. Water, acids, and oxygen are  chemical weathering promoters causing dramatic results over the time.

In the Antarctica, the following conditions would produce chemical weathering:

- Oxidation: is understood by formation of oxides when the rocks react with the environmental gaseous oxygen or the inner oxygen in water.

- Acidification: when oxides such as CO₂, NOₓ or SOₓ react with the water in the atmosphere, their corresponding acids are produced so when it rains they fall over the rocks creating sink holes and caves.

- Hydrolysis: it is caused when water dissolves minerals in a rock, producing different mineral compounds. For instance, feldspar crystals inside the granite react by hydrolysis, forming clay minerals which weakens the rock, making it more likely to break.

Best regards.

andrew11 [14]3 years ago
6 0

Oxidation, Dissolving in acid, Hydrolysis ,Water absorption and Hydration

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Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }  ⇄ \mathtt{3SO_{2(l)}}

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K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}

However, since we are dealing with liquids solutions;

K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}   since the activity of a_{so_3} is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

K = \dfrac{1}{Pso_2Po_2^{1/2}}

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\  \simeq -68 \ kJ/mol

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(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

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\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}

\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K }  \\ \\  K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\  K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }

K =7.98390356\times 10^{11} \\ \\  \mathbf{K = 7.98 \times 10^{11}}

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

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If  Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

Q= \dfrac{1}{Pso_2Po_2^{1/2}}

Since we are dealing with liquids;

Q= \dfrac{1}{1 \times 1^{1/2}}

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

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