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3 years ago

Consider the following reaction at 298 K:

Chemistry

1 answer:

faltersainse [42]3 years ago

8 0

Answer :

$\Delta S_{sys}$ = -1622.8 J/K

$\Delta S_{surr}$ = -94.6 J/K

$\Delta S_{univ}$ = 0 J/K

Explanation : Given,

$\Delta H$ = -483.6 kJ

The given chemical reaction is:

$2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$

First we have to calculate the value of $\Delta S_{sys}$ .

$\Delta S_{sys}=\frac{\Delta H}{T}$

$\Delta S_{sys}=\frac{-483.6kJ}{298K}$

$\Delta S_{sys}=-1.6228kJ/K=-1622.8J/K$

Entropy of system = -1622.8 J/K

As we know that:

Entropy of system = -Entropy of surrounding = 1622.8 J/K

and,

Entropy of universe = Entropy of system + Entropy of surrounding

Entropy of universe = -1622.8 J/K + (1622.8 J/K)

Entropy of universe = 0

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