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Minchanka [31]
3 years ago
14

Find the temperature at which Kp = 42.0 for the reaction

Chemistry
1 answer:
Slav-nsk [51]3 years ago
5 0

Answer: Temperature for the given reaction is 1040 K (approx).

Explanation:

Formula for enthalpy change of a reaction is as follows.

     \Delta H_{rxn} = \Delta H_{products} - \Delta H_{reactants}

For the given reaction equation,

     \Delta H_{rxn} = 2 \times \Delta H_{HI} - (\Delta H_{H_{2}} + \Delta H_{I_{2}})  

Now, putting the given values into the above formula as follows.

    \Delta H_{rxn} = 2 \times \Delta H_{HI} - (\Delta H_{H_{2}} + \Delta H_{I_{2}})  

   \Delta H_{rxn} = 2 \times 25.9 - (0 + 62.26)  

                    = -10460 J/mol

Now, we will calculate the change in its entropy as follows.

           \Delta S = S_{products} - S_{reactants}

                       = 2 \times S_{HI} - (S_{H_{2}} + S_{I_{2}})

                       = 2 \times 206.3 - (131.0 + 260.6)

                       = 21 J/mol

Also, we know that

         \Delta G = RT ln K_{p} = \Delta H_{rxn} - T\Delta S_{rxn}

         -8.314 \times T \times ln(42) = -10460 - T \times 21

                     T = 1036.7 K

                        = 1040 K

Therefore, we can conclude that temperature for the given reaction is 1040 K (approx).

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Calculate the molarity of sodium chloride in a half-normal saline solution (0.45% NaCl). The molar mass of NaCl
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Answer:

0.077 M

Explanation:

Molarity is the representation of the solution.

Molarity:

It is amount of solute in moles per liter of solution and represented by M

Formula used for Molarity

                    M = moles of solute / Liter of solution . . . . . . . . . . (1)

Data Given :

The concentration of half normal (NaCl) saline = 0.45g / 100 g

So,

Volume of Solution =  100 g = 100 mL

Volume of Solution in L =  100 mL / 1000

Volume of Solution =  0.1 L

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Now to find number of moles of Nacl

               no. of moles of NaCl = mass of NaCl / molar mass

               no. of moles of NaCl =  0.45g / 58.44 g/mol

               no. of moles of NaCl =  0.0077 g

Put values in the eq (1)

                  M = moles of solute / Liter of solution . . . . . . . . . . (1)

                  M = 0.0077 g / 0.1 L

                  M = 0.077 M

So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M

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Explanation:

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