1) For each of the following determine if the sign of S for the process. A: delt

Question

1) For each of the following determine if the sign of S for the process.

A: deltaS +

B: deltaS-

- O2(g) --> O2(l)

- 2NO(g) + O2(g) --> 2 N2O(g)

- NaBr(s) --> Na+(aq) + Cl-(aq)

- NH4Cl(s) --> NH3(g) + HCl(g)

2) Arrange the following in order of increasing entropy (low entropy-1 to high entropy-3)

- NaCl (l)

- NaCl (aq)

- NaCl (s)

3) What is the Ssurr in J/molK for a reaction that has a Hsys of 73.1 kJ at 72 oC?

Ssurr  =   Ssys     Ssys =  Hsys /T (pay close attention to signs and units!)

Explanation / Answer

Entropy is the disorder of the system which depends on temperature, physical stateand number of moles etc. As temperature increases, the entropy increases and the entropy of gas> entropy of liquid>entropy of solid for the same substance at the same temperature. More the number of moles more is the disorder/randomness.

Delta S of a process is (S of products - S of reactants)

Q1) O2 (g) ----> O2 (l)

The entropy of liquid is less than the entropy of gas at same temperature.

Thus delta S for this process is negative.

2NO(g) +O2(g) ----> 2N2O(g)

Three gaseous moles are reacting to give 2 gaseous moles of products. As the number of moles decreases , the randomness also decreases, thus delta S is negative.

NaBr(s) ----> Na+(aq) + Br-(aq)

Here the solid is dissolved in water to give ions in aqeuous solution, that is the disorder is increased. Thus delta s is positive.

NH4Cl(s) ----> NH3(g) + HCl(g)

In this reactiona solid substance changed into two moles of gases, that is the entropy is ver much increased. Thus delta S is positive.

Q2) The order is NaCl(s) < NaCl(l) < Nacl(aq)

Q3)

Delta S of system = 73.1x1000J / 345 = 211.88J

Hence Delta S surrounding = -211.88J

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