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AH2298 = 0

A H298 = -165.18 kcal/gmmol of HCO

H2CO3( aq) ^ H+aq) + HCO-(aq) AH298 = +1.82 kcal/gmmol of H2CO3(aq)

A H298 = +165.18 kcal/gmmol of HCO

AH298 = + 288.45 kcal/gmmol of CaCO3(s AH298 = -129.77 kcal/gmmol of Ca2a+q) AH298 = -161.63 kcal/gmmol of CO2—aq)

Example 11.5 A softened municipal water supply enters a residence at 15°C and is heated to 60°C in the water heater. Compare the values of the equilibrium constants for CaCO3 at these two temperatures. If the water was at equilibrium at 25°C, determine if CaCO3 will deposit or not at these two temperatures.

Solution:

= Kt 1 = 4.8( 10-9) in gmol units at 25° C A H298 = -2.95 kcal/gmmol of CaCO3(s) = -2,950 cal/gmmol of CaCO3(s)

Therefore,

2,950

2,950

Thus, the value of equilibrium constant is greater at 60°C than at 15°C. Ans

The value of the equilibrium constant for calcium carbonate at 25°C is 4.8(10 ).

At this condition, the ions Ca and CO3 ions are given to be in equilibrium; thus, will neither deposit nor dissolve CaCO3. At the temperature of 15°C, the value of the equilibrium constant is 4.038(10—9). This value is less than 4.8(10—9) and will require less of the ionized ion; therefore at 15°C, the water is oversaturated and will deposit CaCO3. Ans

At 60°C, the equilibrium constant is 8.10(10—9), which is greater than that at 25°C. Thus, at this temperature, the water is undersaturated and will not deposit CaCO3. Ans

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