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Paul [167]
3 years ago
11

Look at the potential energy diagram for a chemical reaction.

Chemistry
1 answer:
vladimir1956 [14]3 years ago
8 0

Answer: The activation energy is 10 kJ and the reaction is exothermic.

Explanation: Exothermic reactions are those in which heat is released and thus the energy of products is less than the energy of reactants.

Endothermic reactions are those in which heat is absorbed and thus the energy of products is more than the energy of reactants.

Activation energy is the extra amount of energy required by the reactants to cross the energy barrier to convert to products.

Given : Energy of reactants = 40kJ

Energy of activation: (50-40)=10 kJ

Energy of products = 50 kJ

Energy of products = 15 kJ

Thus Energy of products (15kJ) < Energy of reactants(40kJ), the reaction is exothermic as energy has been lost to surroundings in the form of heat.

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A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af
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Explanation:

The given reaction is as follows.

        HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Hence, number of moles of NaOH are as follows.

        n = 0.05 L \times 0.1 M

           = 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

             n = 0.025 L \times 0.1 M

                = 0.0025 mol

According to ICE table,

         HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Initial:     0.005 mol   0.0025 mol              0                  0

Change: -0.0025 mol  -0.0025 mol        +0.0025 mol

Equibm:   0.0025 mol    0                         0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

          [HA] = \frac{0.0025 mol}{V}

        [NaA] = \frac{0.0025 mol}{V}

       [A^{-}] = [NaA] = \frac{0.0025 mol}{V}

Now, we will calculate the pK_{a} value as follows.

          pH = pK_{a} + log \frac{A^{-}}{HA}

       pK_{a} = pH - log \frac{[A^{-}]}{[HA]}

                  = 3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}

                  = 3.42

Thus, we can conclude that pK_{a} of the weak acid is 3.42.

           

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Which of the following is the balanced equation for the
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