Tutorial 6 Notes

Substituting the numbers in 200 100 300 2 mol 600 mol

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substituting the numbers in: ( ) ( ) 2.00 1.00 3.00 2 mol 6.00 mol total A B C D n n n n n x x x x x = + + + = ! + ! + + + = + In the question we are told that at equilibrium there was 0.79 moles of C – this is equal to x and therefore the total number of moles of gas at equilibrium is 6.79. We can use this to calculate the mole fractions for all the compounds at equilibrium: ( ) 2.00 0.79 mol 6.79 mol 0.178 A ! " = = ( ) 1.00 0.79 mol 6.79 mol 0.031 B ! " = = 0.79 mol 6.79 mol 0.116 C ! = = ( ) 3.00 2 0.79 mol 6.79 mol 0.674 D ! + " = = Therefore the mole fractions are 0.178, 0.031, 0.116 and 0.674 for compounds A, B, C and D respectively. (Note that we should not retain too many digits or different values will result. General rule of thumb is that if you are going to use the data for subsequent calculations retain one more than the minimum number of significant figures in the given data (which I assume is 2 from the data of 0.79 moles).) b) What is Kx? This is the equilibrium constant in terms of mole fractions: ( ) ( )( ) ( )( ) 2 2 0.116 0.674 0.178 0.031 9.550 C D A B K ! ! ! ! ! = = = Therefore the mole fraction equilibrium constant is 9.6. This tells us that at equilibrium we will have more products than reactants. c) What is K p ? (The question asks for K – this is the activity based definition of K which we will get into in later sections. For now think of this question as if you had been asked to calculate Kp.) Kp calculations are based on partial pressures. From the notes:

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( ) ( ) 2 2 C D o o A B o o p p p p p p p p p C D o A B K p p p p p = = However, according to Dalton’s law, the partial pressure of a gas in an ideal mixture of gas is equal to the mole fraction times the total pressure or i i p p ! = so that K p becomes: K p = !
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