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emmasim [6.3K]
3 years ago
15

32. Balance the following skeleton equation as illustrated below. (Ref: 282-288)

Chemistry
2 answers:
MA_775_DIABLO [31]3 years ago
8 0

Answer:

1/2H2 (g) + 1/2Cl2 (g) —> HCl (g)

Explanation:

Step 1:

The equation for the reaction is given below:

H2 (g) + Cl2 (g) —> HCl (g)

Step 2:

Balancing the equation.

H2 (g) + Cl2 (g) —> HCl (g)

The above equation can be balanced as follow:

There are 2 atoms of H on the left side and 1 atom on right side. It can be balance by putting 1/2 in front of H as shown below:

1/2H2 (g) + Cl2 (g) —> HCl (g)

There are 2 atoms of Cl on the left side and 1 atom on the right side. It can be balance by putting 1/2 in front of Cl as shown below:

1/2H2 (g) + 1/2Cl2 (g) —> HCl (g)

Now we can see that the equation is balanced

aliina [53]3 years ago
5 0

Answer: H2 + Cl2 => 2HCl

Explanation: the equation is now balanced. 2 H and 2 Cl atoms both in the reactant and product side.

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A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch
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Answer:

The reactions free energy \Delta G = -49.36 kJ

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      The pressure of (NO) is P_{NO} = 9.20 \ atm

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              2NO_{(g)} + Cl_2 (g)    ⇆   2 NOCl_{(g)}

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                    \Delta G^o = 2 \Delta G^o _{NOCl} -   \Delta G^o _{Cl_2}  - 2 \Delta G^o _{NO}

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                 \Delta G^o _{NO} = 86.55 kJ/mol

 The Standard state  free energy for Cl_2 is  constant with a value                  

             \Delta G^o _{Cl_2} = 0kJ/mol

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         \Delta G^o _{NOCl} =66.1kJ/mol

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        \Delta G^o = 2 * 66.1 - 0 - 2 * 87.6

                = -43 kJ/mol

The pressure constant is evaluated as

         Q =  \frac{Pressure \ of  \ product }{ Pressure  \ of \ reactant }

Substituting  values  

        Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}

           = 0.0765

The free energy for this reaction is evaluated as

           \Delta  G  =  \Delta  G^o  + RT ln Q

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          T is temperature in K  with a given value of  T = 25+273 = 298 K

   Substituting value

                \Delta  G  = -43 *10^{3} + 8.314 *298 * ln [0.0765]

                       = -43-6.36

                      \Delta G = -49.36 kJ

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