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Solution: Consider the following reaction at 298 K:2 H2 (g) + O2 (g) → 2 H2O (g)     ΔH= -483.6 KJCalculate the following quantities.i) ΔSsys=ii) ΔSsurr=iii) ΔSuniv=


Consider the following reaction at 298 K:

2 H(g) + O(g) → 2 H2O (g)     ΔH= -483.6 KJ

Calculate the following quantities.

i) ΔSsys=

ii) ΔSsurr=

iii) ΔSuniv=


In order to find the change in entropy for the system, we need values for S of the product and both reactants.

$S\ of\ H_2 = 130.7\ \frac{Jmol}{K}\\
S\ of\ H_2O = 188.8\ \frac{Jmol}{K}\\
S\ of\ O_2 = 205.2\ \frac{Jmol}{K}$

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