Example 5.8. Caculating the temperature the reaction startes

Problem: Using the data and calculations from previous problem, calculate the temperature at which decomposition of NaHCO_3(s) becomes spontaneous. Suppose H and S to be temperature independent.

Solution: the _reaction is unfavorable for enthalpy (H  0), but favorable for entropy (S  0). Therefore, G for this reaction will be negative at high temperature. To determine the temperature at which decomposition of NaHCO_3(s) starts, we should substitute the values of H°_r and S°_r into the equation

G = H  TS,

suppose G = 0 and solve for T.

G = H  TS = 0

T = H/S = 128/0.228 = 561.4 K (288°C)

Answer: decomposition of NaHCO_3(s) becomes spontaneous at 288°C.

1 The enthalpy of a system equals to the sum of the internal energy of a system (U) and a product of the system’s pressure and volume

H = U + PV

2 The enthalpy change equals to the change of the internal energy of substances plus expansion work

ΔH = ΔU + PΔV

3 Otherwise a perpetuum mobile (device that creates energy) could be done.

4 The first low of thermodynamics is a form of the law of conservation of matter. It states that energy can not be created or destroyed; it can only be redistributed or changed from one form to another. The change in internal energy of a system is a sum of energy released or absorbed and the work done.

5 Entropy of a system can be calculated using Boltzmann’s equation:

S = k ln w

In this equation, k is a constant (equal to R/N_A) and w is the number of microstates describing the macrostate of the system. (The greater is the number microstates describing the macrostate, the greater is the entropy of a system.)

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