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adelina 88 [10]
3 years ago
14

Please help me with question 7. Thank you so much.

Chemistry
1 answer:
tigry1 [53]3 years ago
4 0

Answer:

The system is not at equilibrium and the reaction will proceed to the left.

Explanation:

Step 1: Write the balanced equation

H₂(g) + CO₂(g) ⇄ CO(g) + H₂O(g)

Step 2: Calculate the reaction quotient (Q)

The reaction is calculated in the same way as the equilibrium constant (Kc) but it uses the concentrations at any time.

Q = [CO] × [H₂O] / [H₂] × [CO₂]

Q = 0.610 × 0.695 / 0.425 × 0.500 = 2.00

Since Q ≠ Kc, the reaction is not at equilibrium.

Since Q > Kc, the reaction will proceed to the left.

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- 622.5kJ

Explanation:

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-  3 (N2O  + 3H2     --->  N2H4 + H2O    ΔH° 2 = - 317 kJ)

2)      2NH3 + 3N2O ----> 4N2 + 3H2O        ΔH° 1= - 1010kJ

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2NH3  + 3N2H4 --->4N2+9H2,     ΔH° 1 -  3*ΔH° 2

2NH3  + 3N2H4  --->4N2+9H2,        ΔH° 1 -  3*ΔH° 2

3) -(2NH3 +1/2O2 ---> N2H4 + H2O,     ΔH°3 = -143 kJ)

   -2NH3 - 1/2O2 ---> - N2H4 - H2O,    - ΔH°3 = 143 kJ

 2NH3  + 3N2H4 --->4N2+9H2,        ΔH° 1 -  3*ΔH° 2

<u>  -2NH3 - 1/2O2              ---> - N2H4 - H2O,             - ΔH°3 = 143 kJ</u>

 4N2H4 + H2O  ---> 4N2  + 9H2 + 1/2O2,    ΔH° 1 -  3*ΔH° 2 - ΔH°3

4)

H2 + 1/2O2 ---> H2O,    ΔH° 4 = - 286 kJ

9*(H2 + 1/2O2 ---> H2O,    ΔH° 4 = - 286 kJ)

9H2+ 9/2 O2 ---> 9H2O,  9*ΔH° 4 = 9*(- 286) kJ

4N2H4 + H2O  ---> 4N2  + 9H2 + 1/2O2,    ΔH° 1 -  3*ΔH° 2 - ΔH°3

<u>9H2+ 9/2 O2    ---> 9H2O,                             9*ΔH° 4 = 9*(- 286) kJ</u>

4N2H4 +4O2 --->4N2+8H2O,         ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

5)

1/4*(4N2H4 +4O2 --->4N2+8H2O,    ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

N2H4 +O2 --->N2+2H2O,        1/4( ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)

6)

N2H4 +O2 --->N2+2,        

1/4( ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)=

=1/4(-1010kJ - 3*(-317kJ) - (-143kJ) + 9*(-286kJ))= - 622.5 kJ

   

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