Question

19 Question (1point) e See page 658 Automobiles and trucks pollute the air with NO. At 2000.0°C, Kc for the reaction 2NO (g) N2 ()+02(g) At 2000.0°C Kc= 4.100 x 10-4and AH° = 180.6 kJ. 3rd attempt See Periodic Table See Hint What is the value of Kc at 63.00°C? x 10-8 2.337

Answer 1

Answer:-

Given:-

first temperature (T₁) = 2000.0 ⁰C = 2000.0 + 273 = 2273.0 K

second temperature (T₂) = 63.0 ⁰C = 63.0 + 273 = 336 K

enthalpy of reaction ($\Delta$H) = 180.6 kJ = 180600 J

equilibrium constant (K_c1) at 2000.0 ⁰C = 4.100 $\times$ 10^-4

equilibrium constant (K_c2) at 63.0 ⁰C = ?

As we know that

gas constant (R) = 8.314 JK^-1mol^-1

So

According to the integrated form of Vant Hoff equation

log (K₂ / K₁) = $\Delta$H / 2.303 R [T₂ - T₁ / T₁T₂]

therefore

log (K_c2 / K_c1) = $\Delta$H / 2.303 R [T₂ - T₁ / T₁T₂]

log (K_c2 / 4.100 $\times$ 10^-4 ) = 180600 / 2.303 $\times$ 8.314 [336 - 2273.0 / 2273.0 $\times$ 336 ]

log (K_c2 / 4.100 $\times$ 10^-4 ) = 180600 / 19.147142‬ [ - 1937.0 / 763728‬  ]

log (K_c2 / 4.100 $\times$ 10^-4 ) = 9432.22 $\times$ [ - 1937.0 / 763728‬ ]

log (K_c2 / 4.100 $\times$ 10^-4 ) = 9432.22 $\times$ - 1937.0 / 763728‬

log (K_c2 / 4.100 $\times$ 10^-4 ) = -18270210.14‬ / 763728‬

log (K_c2 / 4.100 $\times$ 10^-4 ) = - 23.9224

On taking antilog of the above value

K_c2 / 4.100 $\times$ 10^-4 = 0.119564 $\times$ 10^-23

K_c2 = 4.100 $\times$ 10^-4$\times$ 0.119564 $\times$ 10^-23​​​​​​​

equilibrium constant (K_c2) at 63.0 ⁰C = 0.4902 $\times$ 10^-27 (i.e the answer)

or

equilibrium constant (K_c2) at 63.0 ⁰C = 4.902 $\times$ 10^-28 (i.e the answer)