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For the reaction below G° = + 33.0 kJ, H° = + 92.2 kJ, and S° = + 198.7 J/K. Estimate the temperature (K) at which this reaction becomes spontaneous. 2 NH3(g) → N2(g) + 3 H2(g)
a) 0.464k b)166k c)214k d)298k e)464k

Respuesta :

Answer:

D

Explanation:

A reaction becomes spontaneous when the value of the change in free energy is negative.

The relationship between the enthalpy, the entropy and the free energy in a chemical reaction is given below:

G = H – TS

Now, let’s input the values that we have.

One important thing to note is that we have to convert the KJ to Joules and this mean we have to multiply by a conversion factor of 1000. This mean 33KJ is 33,000J

33,000 = 92,200 - 198.7T

33,000 - 92,200 = -198.7T

-59,200 = - 198.7T

T = 59,200/198.7

T = 298k

Answer:

464K

Explanation:

△G° = △H° -T△S° ,

33 = 92.2 - T(198.7/1000)  [divide by 1000 to convert J to kJ]

T = (92.2 - 33)/(198.7/1000) = 298K

If reaction is spontaneous, △G°  has to be negative, so

0 = △H° -T△S°

T = △H°/△S° = 92.2/(198.7/1000) = 464K

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