Question

Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures are, respectively, 1.00 atm and 0.500 atm. When the system comes to equilibrium at 1100 K, the total pressure in the flask is found to be 1.35 atm. Given: 2SO2(g) + O2(g) ⇌ 2SO3(g); ΔrH = − 198.2 kJ. mol-1 1.1 Calculate Kp at 1100 K

Answer 1

Answer:

The answer is "$\bold{0.525\ \ atm^{-1}}$"

Explanation:

Given equation:

$2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$

Given value:

$\Delta rH =-198.2 \ \ \frac{KJ}{mol}$

$Kp=1100 \ K$

$\Delta x = 2-(2+1)$

     $= 2-(2+1)\\\\= 2-(3)\\\\= -1$

$\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right$

calculating the total pressure on equilibrium=  $(1-2x)+(0.5-x)+2x \ atm$

                                                                         $= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x$

$\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15$

calculating the pressure in  $So_2$:

$= (1-2 \times 0.15)$

$= 1-0.30 \\\\ =0.70 \ atm$

calculating the pressure in  $O_2$:

$= (0.5- 0.15)\\\\= 0.35 \ atm$

calculating the pressure in  $So_3$:

$= (2 \times 0.15)\\\\= (.30) \ atm$

Calculating the Kp at 1100 K:

$= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1}$