Question

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

ΔS > 0; ΔS < 0; too close to decide
I2(g) + Cl2(g) -----> 2ICl(g)
2NOBr(g) -------> 2NO(g) + Br2(g)
CO2(g) + H2(g) -----> CO(g) + H2O(g)
2H2O2(I) ------> 2H2O(I) + O2(g)

Answer 1

Answer: a)
`I_2(g)+Cl_2(g)\rightarrow 2ICl(g)`: too close to decide.

b)
`2NOBr(g)\rightarrow 2NO(g)+Br_2(g)`:
`\Delta S` > 0.

c)
`CO_2(g)+H_2(g)\rightarrow CO(g)+H_2O(g)`: too close to decide.

d)
`2H_2O_2(l)\rightarrow 2H_2O(l)+O_2(g)`:
`\Delta S` > 0.

Step-by-step explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

For the reaction:

a)
`I_2(g)+Cl_2(g)\rightarrow 2ICl(g)`

In this reaction 2 moles of gaseous reactants are converting to 2 moles of gaseous products. Thus
`\Delta S` is too close to decide.

b)
`2NOBr(g)\rightarrow 2NO(g)+Br_2(g)`

In this reaction 2 moles gaseous reactants is getting converted to 3 moles of gaseous products. Thus the randomness will increase and hence entropy will also increase.Thus
`\Delta S` > 0.

c)
`CO_2(g)+H_2(g)\rightarrow CO(g)+H_2O(g)`

In this reaction 2 moles of gaseous reactants are converting to 2 moles of gaseous products. Thus
`\Delta S` is too close to decide.

d)
`2H_2O_2(l)\rightarrow 2H_2O(l)+O_2(g)`

In this reaction 2 moles liquid reactants is getting converted to 2 moles of liquid and 1 mole of gaseous products. Thus the randomness will increase and hence entropy will also increase.Thus
`\Delta S` > 0.