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professor190 [17]
3 years ago
14

Use Le Châtelier's principle to predict how the equilibrium for the weak base methylamine responds to the indicated changes. CH

3 NH 2 ( aq ) + H 2 O ( l ) − ⇀ ↽ − CH 3 NH + 3 ( aq ) + OH − ( aq ) Δ H ∘ rxn = 103 kJ / mol
Chemistry
1 answer:
Anarel [89]3 years ago
3 0

The question is incomplete, here is the complete question:

Use Le Châtelier's principle to predict how the equilibrium for the weak base methylamine responds to the indicated changes.

CH_3NH_2(aq.)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq.)+OH^-(aq.);\Delta H^o_{rxn}=103kJ/mol

1) adding KOH

2) Increasing the temperature

<u>Answer:</u>

<u>For 1:</u> The reaction will shift in left direction that is towards the reactants.

<u>For 2:</u> The reaction will shift in right direction that is towards the product.

<u>Explanation:</u>

For the given chemical equation:

CH_3NH_2(aq.)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq.)+OH^-(aq);\Delta H^o_{rxn}=103kJ/mol

We are given:

\Delta H^o_{rxn}=103kJ/mol

As, enthalpy of the reaction is positive. So, it is an endothermic reaction.

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

A+\text{heat}\rightleftharpoons B

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

  • <u>For 1:</u> Adding KOH

As, hydroxide ions are present on the product side of the reaction. So, when more hydroxide ion is added, the concentration of hydroxide ion increases on the product side.

According to Le-Chatelier's principle, to minimize the effect, the equilibrium will shift in the direction where hydroxide ion concentration decreases, which is to the reactant side.

Hence, the reaction will shift in left direction that is towards the reactants.

  • <u>For 2:</u> Increasing the temperature

As, heat is getting absorbed during a chemical reaction. This means that temperature is getting increased on the reactant side. If the temperature in the equilibrium is increased, the equilibrium will shift in the direction where, temperature is getting decreased.

Hence, the reaction will shift in right direction that is towards the product.

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\begin{gathered} n=\frac{0.994737\cancel{\text{atm}}\times0.0425\cancel{L}}{0.0821\frac{\cancel{L}\cdot\cancel{\text{atm}}}{\text{mole}\cdot\cancel{K}}\times299\cancel{K}} \\ n=\frac{0.994737\times0.0425}{0.0821\times299} \\ n=\frac{0.0422763225}{24.5479} \\ n=0.00172\text{moles} \end{gathered}

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\begin{gathered} \text{Mass}=0.00172\cancel{\text{moles}}\times\frac{110.262g}{\cancel{\text{mole}}} \\ \text{Mass}=0.1899\text{grams} \end{gathered}

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