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lidiya [134]
4 years ago
8

If ammonia is manufactured at 356 k, is the reaction spontaneous, given that the enthalpy and entropy change for the reaction ar

e -93 kj/mol and -198 j/mol k, respectively?
Chemistry
1 answer:
Alchen [17]4 years ago
5 0
The reaction will be spontaneous if Gibb free energy is negative, according the following relation:  
G = H - T*S  
where G is Gibbs free energy, T is change of enthalpy , T is temperature and S is change of entropy  
In the case of ammonia:  
G = -93000 - 356*(-198) = -93000 + 70488 = -22512 j/mol  
As G < 0 then the reaction is spontaneous.
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The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sampl
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Answer: Concentration of NH_3 in the equilibrium mixture is 0.31 M

Explanation:

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The given balanced equilibrium reaction is,

                 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

Initial conc.            x                0           0

At eqm. conc.     (x-2y) M     (y) M   (3y) M

The expression for equilibrium constant for this reaction will be:

3y = 0.729 M

y = 0.243 M

K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}

Now put all the given values in this expression, we get :

K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}

0.967=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}

x=0.80

concentration of NH_3 in the equilibrium mixture = 0.80-2\times 0.243=0.31

Thus concentration of NH_3 in the equilibrium mixture is 0.31 M

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3 years ago
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Answer:

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Answer:

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