Question #54165

2. a. How will you decide whether a process is spontaneous in this experiment?
b. If you were to use this method after you had observed the fate of an ice cube at
25’C, what would you conclude about the spontaneity of the following process?
Why?
H20(s)  H20(l)
c. The standard enthalpy change for this process is 6.01 kJ/mol. What is the
minimum value for the standard entropy change, based upon your conclusions
about the spontaneity?
3. During this experiment, you will be required to prepare 100 mL of a 1.0 M solution of
NaNO3. Calculate the mass of NaNO3 (to the nearest tenth of a gram) that will be
required?

Expert's answer

Answer on Question #54165 – Chemistry – Other

Task:

2. a. How will you decide whether a process is spontaneous in this experiment?

b. If you were to use this method after you had observed the fate of an ice cube at 25C25^{\circ}\mathrm{C}, what would you conclude about the spontaneity of the following process?

Why? H2O(s)H2O(l)\mathrm{H}_2\mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{l})

c. The standard enthalpy change for this process is 6.01kJ/mol6.01 \, \mathrm{kJ/mol}. What is the minimum value for the standard entropy change, based upon your conclusions about the spontaneity?

3. During this experiment, you will be required to prepare 100mL100\,\mathrm{mL} of a 1.0M1.0\,\mathrm{M} solution of NaNO3\mathrm{NaNO}_3.

Calculate the mass of NaNO3\mathrm{NaNO}_3 (to the nearest tenth of a gram) that will be required?

Answer:

2 a. The process will be spontaneous if Gibbs energy is ΔG<0\Delta G < 0.

2 b. Fate of an ice cube at 25C25^{\circ}\mathrm{C} is a spontaneous process.

2 c. To be spontaneous the process must fulfill the condition ΔHTΔS<0\Delta H - T\Delta S < 0.

6.01(25+273)ΔS<06.01 - (25 + 273) \cdot \Delta S < 0

ΔS>0.02\Delta S > 0.02

3.


CM=vVv=CMVC_M = \frac{v}{V} \quad v = C_M \cdot Vv=mMm=vMv = \frac{m}{M} \quad m = v \cdot Mv(NaNO3)=0.11.0=0.1molv(\mathrm{NaNO}_3) = 0.1 \cdot 1.0 = 0.1\,\mathrm{mol}M(NaNO3)=85g/molM(\mathrm{NaNO}_3) = 85\,\mathrm{g/mol}m(NaNO3)=0.185=8.5gm(\mathrm{NaNO}_3) = 0.1 \cdot 85 = 8.5\,\mathrm{g}


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