Tutorial 6 Notes - CHM2132 - Tutorial Notes for Oct. 20th,...

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CHM2132 - Tutorial Notes for Oct. 20 th , 23 rd 1) Atkins & dePaula, Problem 3.7 We need to determine the entropy of NH 3 (g) at 100 o C. We can find the standard entropy of NH 3 (g) at 298 K from Table 2.6 in the textbook - all we have to calculate is the amount of entropy that arises from increasing the temperature of this gas by an additional 75 o C: ! S m o 373 K ( ) = ! S m o 298 K ( ) + ! S m o 298 K " 373 K ( ) where ! S m o 298 K " 373 K ( ) is the entropy change of raising the temperature. In order to calculate the second entropy term we must start from the general equation for entropy: ! S = " q rev T # As this is a constant pressure heating process we can start with the general equation for constant pressure heat capacity to find the reversible heat: ! H ! T " # $ % p = C p , m Rearranging into a form to isolate for enthalpy (which is the reversible heat): dH = C p , m dT = ! q rev However, in this case we are told that the heat capacity is temperature dependent according to: C p , m = a + bT + c T 2 Therefore we must substitute this equation into the equation for reversible heat: q rev = a + bT + c T 2 " # $ % dT Now this expression for reversible heat can be substituted back into the equation for entropy: ! S = C p , m dT T " = a + bT + c T 2 # $ % ( T " dT = a T + b + c T 3 # $ % ( " dT Now using the following general integration formulae: x n ! dx = 1 n + 1 x n + 1 + c and dx x ! = ln x ( ) + c We can solve for entropy:
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! S = a ln T + bT " c 2 T 2 T 1 T 2 = a ln T 2 T 1 # $ % ( + b ! T " c 2 1 T 2 2 " 1 T 1 2 # $ % ( Now we can find the values for a , b and c for NH 3 (g) and plug into this equation. However, since we know the absolute entropy at 298 K we will only need to use this
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Tutorial 6 Notes - CHM2132 - Tutorial Notes for Oct. 20th,...

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