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vichka [17]
4 years ago
7

If δh = -70.0 kj and δs = -0.300 kj/k , the reaction is spontaneous below a certain temperature. calculate that temperature.

Chemistry
2 answers:
IgorC [24]4 years ago
5 0
The  temperature  for  the  reaction  is  calculated  as  follows

delta  H/ delta S


delta  H=-70 Kj
delta  S  =-0.300  kj/k
temperature  is  therefore = -70kj/-0.300kj/k  =233.33 K
Dafna11 [192]4 years ago
5 0

Answer: The reaction is spontaneous below a temperature of 233.3 K.

Explanation:

Using Gibbs Helmholtz equation:

\Delta G=\Delta H-T\Delta S

\Delta G = Gibbs free energy  

\Delta H = enthalpy change  = -70 kJ

\Delta S = entropy change  = -0.300 kJ/K

T = temperature in Kelvin

\Delta G= +ve, reaction is non spontaneous

\Delta G= -ve, reaction is spontaneous

\Delta G= 0, reaction is in equilibrium

At equilibrium : T\Delta S=\Delta H

T=\frac{\Delta H}{\Delta S}

T=\frac{-70.0kJ}{-0.300kJ/K}

T=233.3K

Thus the reaction is spontaneous below a temperature of 233.3 K.

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