2 x_ 6(g) 4 x_ 3(g) a 20 L box containing X_6 and X_3 at equilibrium at 1500 K . K_p = 4 10^ 18 atm . Assuming P_ x_3 > > P_ x_6 and total pressure is 10 atm .The partial pressure of X_6 will be

2 x_ 6(g) 4 x_ 3(g) a 20 L box containing X_6 and X_3 at equilibr…

MCQ

$2{x_{6(g)}} \rightleftharpoons 4{x_{3(g)}}$

a $20\ L$ box containing $X_6$ and $X_3$ at equilibrium at $1500\ K$ . $K_p = 4 \times 10^{18}\ atm$ . Assuming ${P_{{x_3}}} > > {P_{{x_6}}}$ and total pressure is $10\ atm$ .The partial pressure of $X_6$ will be

✓
$5 \times {10^{ - 8}}\ atm$
B
$5 \times {10^{ - 6}}\ atm$
C
$2 \times {10^{ - 8}}\ atm$
D
None

✓

Answer

Correct option: A.

$5 \times {10^{ - 8}}\ atm$

a
${{\text{K}}_{\text{p}}} = \left( {\frac{{{X_3}}}{{{X_6}}}} \right) = 4 \times {10^{18}} = \frac{{{{(10)}^4}}}{{{{({X_6})}^2}}}$

${\left( {{{\text{x}}_6}} \right)^2} = \frac{{{{10}^4}}}{{4 \times {{10}^{18}}}} = 25 \times {10^{ - 16}}$

$\mathrm{X}_{\mathrm{6}}=5 \times 10^{-8}\, \mathrm{atm}$

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