Step 4: Calculate the partial pressures of A, B, C, and D P A = x A P total ; P B = x B P total ; P C = x C P total ; P D = x D P total ; 2 θ θ

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1 1. sealed container was filled with 1.00 mol,.00 mol, and 1.00 mol D. The following reaction is expected to occur (g) + (g) C(g) + D(g). When equilirium state is estalished, the resulting mixture contains 0.6 mol C. The final pressure is 1.00 ar. Calculate (a) the equilirium constant., and () the standard reaction Gis energy. (a) Step 1: calculate the changes of,, and D from C. nowing C 0.6 mol So - 1*(0.6/) -0. mol - *(0.6/) mol D *(0.6/) 0.4 mol Step : calculate the amounts of,, C, and D at the equilirium state n mol n mol n C 0.6 mol n D mol Step : Calculate mole fractions of,, C, D x n /(n + n + n C + n D ) 0.8/4.4 x n /(n + n + n C + n D ) 1.6/4.4 x C n C /(n + n + n C + n D ) 0.6/4.4 x D n D /(n + n + n C + n D ) 1.4/4.4 (Remark: you do not need to complete the numerical calculations here in order to reduce the round-off error) Step 4: Calculate the partial pressures of,, C, and D P x *P total ; P x *P total ; P C x C *P total ; P D x D *P total ; Step 5: ssuming that gases are perfect, α j P j /P ө Pc PD ( ) ( ) θ θ P P P P ( ) ( ) θ θ P P plug in these numers with P total 1 ar, ecause the amount of C has only one significant figure, the final result should e: 0.01

2 () r G ө - RTln J -1 mol -1 * 98,15 * ln (0.0107) Jmol -1 1 x 10 4 Jmol -1. In the presence of metallic catalyst, nitrogen and hydrogen react to form ammonia in a process called the Haer process. The Haer reaction is N (g) + H (g) NH (g) (a) Using the data in ppendix (tkin s textook), compute p for the Haer reaction at () Computer the equilirium constant for the reaction at 500. In this calculation, you may assume that H o is a constant equal to its value at (c) t what temperature would we have p 1? You may again assume that is a constant. (a) The equilirium constant is given y: θ rg ln RT Using the data in tale.6, we otain r G ө f G ө (NH ) - f G ө (N ) f G ө (H ) *( kj mol -1 ) -.90 kj mol -1 Therefore.90kJ mol ln 8.145J mol ln x 10 5 () Since r H ө is a constant, using the solution of the van t Hoff equation ln ln 1 - ( r H ө /R)(1/T 1/T 1 ) (1) The enthalpy change in the reaction: r H ө f H ө (NH ) f H ө (N ) f H ө (H ) *(-46.11)kJ mol -1 * kj mol -1 sustituting into the aove equation (1) gives ln( ) ln(5.81*10 5 ) - (-9. kj mol -1 /8.14J -1 mol -1 )(1/500 1/98.15) ln( ) ln(5.81*10 5 )

3 (c) Using the same equation as () ln ln 1 - ( r H ө /R)(1/T 1/T 1 ) It is given that 1 at temperature T, sustituting values into the aove equation, We get ln(1.0) ln(5.81*10 5 ) - (-9. kj mol -1 /8.14J -1 mol -1 )(1/T 1/98.15) so 1/T T 46.5 Since 1 has only one significant figure, the final result should read as : T 5 x10. The equilirium constant p for the Haer reaction at 400 is Suppose that moles of H (g) and 1 mole of N (g) are mixed at 400 in a closed, fixed-volume liter container. When equilirium is reached, what are the partial pressures of H (g), N (g), and NH (g)? What is the percent conversion to NH (g)? (ssume that all gases are ideal) Write and alance the Haer reaction N (g) + H (g) NH (g) ssume that the numer of moles of NH present at equilirium equals x. In this case the stoichiometry of the reaction gives: nh (g) at equilirium.0 *(x/) and nn (g) at equilirium 1.0 (x/) The partial pressures are: nrt ( x)( lar mol )(400) P H ( g ) L (1-x) ar, nrt (1 x)( lar mol )(400) P N ( g ) L (1-x) ar and nrt x( lar mol )(400) PNH g ) ( L.576x ar Sustituting into the expression for p p P NH ( g ) P N ( g ) ( P P ) θ H ( g) (.576x) * (1 0x) * (1 x) 0.165

4 Therefore: x (1 x) x (1 x) or ( no physical meaning as x cannot e negative) x.05698x so x 1.69 or x 1.69 can e dropped ecause it leads to a result without physical meaning Using the value x nrt P H ( g ) 10.6 ar; nrt P N ( g ).5 ar nrt PNH ( g ) 6. ar () the percentage conversion can e defined in several ways: % conversion the amount of the theoretical amount of NH NH produced 100% produced when all N is consumed Or Or the amount of N reacted % conversion the initial amount of N the amount of H reacted % conversion the initial amount of H x 100% x 100% so if we use N as our measure %conversion x/1.0mol * 100% 78.8% 4. State the direction of the equilirium shift in each of the following cases, and stipulate whether the shift is taking place ecause the equilirium constant is increasing or decreasing or ecause of the necessity to maintain the equilirium constant s value unchanged: (a) In the reaction NO (g) N O 4 (g) + 57.kJ, the total pressure is decreased.

5 () In the aove reaction, the temperature is decreased. (c) In the aove reaction, some of the N O 4 (g) is removed (a) The equilirium will shift toward the reactants to increase the numer of moles of gas, therey relieving the stress produced y the pressure reduction. Thermodynamically, this shift occurs in such manner to maintain a constant value for, which would otherwise increase. () The equilirium shifts towards the products to produce more heat and therefore reduces the disturance caused y the removal of heat. Thermodynamically, we have H < 0, so that must increase as T decreases. Therefore, the value of rises and the equilirium shifts toward the product side. (c) The equilirium will shift toward the products to increase the numer of moles of N O 4 (g), therefore reduces the stress produced y reducing N O 4 gas. Thermodynamically, this shift occurs in such manner to maintain a constant value for, which would otherwise decrease. 5. Derive an exact expression for ν / for the titration of an aqueous solution of a weak ase with an aqueous solution of a strong acid. To solve this prolem, we are going to need the following five conditions are: (a) Electrical neutrality: [H + + [H O + [ - + [OH - () Conservation of groups: [ + [H Where is the initial volume of ase, 0 is the initial concentration of the weak ase, and is the volume of titrant (acid) added. (c) Concentration of - groups: [ Where 0 is the initial concentration of the strong acid. + [ H [ OH (d) Protonation equilirium of : [ (e) utoprotolysis equilirium: w [H O + [OH - First we express condition () in terms of [H + and [OH - y using condition (d) to eliminate [ [H + ( + )( [ OH + ) 0

6 Next we use this relation and condition (c) and condition (e) to eliminate [H O + ( + )([ OH + ) + [ OH w + [ OH Now use the given condition *v to replace v ( + v)( [ OH + ) + [ OH w + [ OH v so v ( 1 + v)( [ OH + ) + [ OH v 0 w + [ OH reorganize the aove equation to get ([ OH + 0 [ OH w )([ OH + ) v ([ OH + )( [ OH ) + [ OH w 0

This is important to know that the P total is different from the initial pressure (1bar) because of the production of extra molecules!!! = 0.

This is important to know that the P total is different from the initial pressure (1bar) because of the production of extra molecules!!! = 0. Question 1: N O 4 (g) NO (g) Amounts of material at the initial state: n 0 0 Amounts of material at equilibrium: (1 α)n 0 αn 0 Where α equals 0.0488 at 300 K and α equals 0.141 at 400K. Step 1: Calculate