At712 C the concentration equilibrium constant Ke48 x 105 fo

At-7.12 °C the concentration equilibrium constant Ke-4.8 x 105 for a certain reaction. Here are some facts about the reaction: . If the reaction is run at constant pressure, 102. kJ/mol of heat are absorbed. . The net change in moles of gases is-1. The constant pressure molar heat capacity C,- 1.50 J-mol K n Yes. x10 Using these facts, can you calculate Keat-16. oC? No. If you said yes, then enter your answer at right. Round it to 2 significant digits. Yes, and K will be Explanation

Solution

Sol:-

(a) Ans :- Yes , we can calculate equilibrium constant ( say K_c2) at - 16 ⁰C by using vant\'s hoff equation .

i.e

log K_c2 / K_c1 = deltaH / 2.303 R [ T₂ - T₁ / T₁ T₂ ]

(b) Calculation of K_c2 :-

log K_c2 - log K_c1 = deltaH / 2.303 R [ T₂ - T₁ / T₁ T₂ ] ..............(1)

here K_c2 and K_c1  are the equilibrium constants at temperatures T₁ and T₂ respectively .

R = Gas constant = 8.314 x 10^-3 KJ/K mol

Now given that K_c1 = 4.8 x 10⁵ , T₁ = - 7.12 ⁰C = 266.03 K

K_c2 = ?, T₂ = - 16 ⁰C = 257.15 K

delta H = heat change at constant pressure = 102 KJ/mol

Now substitute these values in equation (1) , we have

log K_c2 - log 4.8 x 10⁵ = 102 KJ mol^-1 / 2.303 x 8.314 x 10^-3 KJ K^-1 mol^-1 [ 257.15 K - 266.03 K / 257.15 K x 266.03 K ]

log K_c2 - 5.68  = 5327.166   [ -8.88 / 68409.6 ]

log K_c2 - 5.68  = 5327.166   [ - 0.0001298 ]

log K_c2 - 5.68  = - 0.6915

log K_c2 = - 0.6915 + 5.68

log K_c2 = 4.988

K_c2 = 10^4.988

K_c2 = 9.7 x 10⁴

Hence value of concentration equilibrium constant i.e K_c2 at - 16⁰C is equal to 9.7 x 10⁴ .

(c)

K_c2 at - 16 ⁰C ( i.e 9.7 x 10⁴) is smaller than K_c1 at - 7.12 ⁰C ( i.e 4.8 x 10⁵) , because the given reaction is endotermic in the forward direction ( as given delta H = + 102 KJ/mol ) therefore for endothermic reaction decrease in temperature ( i.e from - 7.12 ⁰C to - 16 ⁰C ) will decrease the value of K_c .

Hence K_c2 < K_c1 .