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Predict whether S for each reaction would be greater than zero, less than zero, or too close to zero to decide.

a. H2(g)+Cl2(g)→2HCl(g)
b. 2H2O2(l)→2H2O(l)+O2(g)
c. CO(g)+3H2(g)→CH4(g)+H2O(g)
d. CH4(g)+2O2(g)→CO2(g)+2H2O(g)
e, 2SO3(g)→2SO2(g)+O2(g)

Sagot :

Jufsanabriasaidn

Answer:

a. Too close to zero.

b. Higher than zero.

c. Lower than zero.

d. Too close to zero.

e. Higher than zero.

Explanation:

The entropy of a reaction, S, is positive when the disorder of the system increases. That is:

Solid → Liquid → Gas S>0

When S<0, the disorder decreases:

Gas → Liquid → Solid S<0

Thus:

a. H2(g)+Cl2(g)→2HCl(g)

2 moles of gas produce 2 moles of gas:

S too close to zero.

b. 2H2O2(l)→2H2O(l)+O2(g)

A liquid is producing a gas:

S>0

c. CO(g)+3H2(g)→CH4(g)+H2O(g)

4 moles of gas produce 2 moles of gas, the disorder decreases:

S<0

d. CH4(g)+2O2(g)→CO2(g)+2H2O(g)

3 moles of gas produce 3 moles of gas:

S too close to zero.

e, 2SO3(g)→2SO2(g)+O2(g)

2 moles of gas produce 3 moles of gas.

S>0

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