Question

Automobiles and trucks pollute the air with NO. At 2000.0°C, Kc for the reaction At 2000.0°C Kc = 4.100 × 10–4 and ΔH° = 180.6 kJ
.What is the value of Kc at 81.00°C?

Answer 1

The value of $K_c$ at 81.00°C is $6.972\times 10^{-14}$

Explanation:

According to the vant hoff equation,

$K=A\times e^{\frac{-\Delta H}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{\Delta H}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = equilibrium constant at $2000.0^oC$ = $4.100\times 10^{-4}$

$K_2$ = equilibrium constant at $81.00^oC$ =?

$\Delta H$ =enthalpy for the reaction = 180.6 kJ = 180600 J

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $2000.0^oC=273+2000.0=2273.0K$

$T_2$ = final temperature = $81.00^oC=273+81.00=354.00K$

Now put all the given values in this formula, we get

$\log (\frac{K_2}{4.100\times 10^{-4}})=\frac{180600J}{2.303\times 8.314J/moleK}[\frac{1}{2273.0K}-\frac{1}{354.00K}]$

$K_2=6.972\times 10^{-14}$

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