NO and O₂ react to form NO₂ at 298K according to the reaction.
${NO}_{(g)} + \frac{1}{2} O_{2 (g)} ⇌ {NO}_{2 (g)}$
Initially pressure of NO and O₂ are 1 atm and 11/4 atm respectively and the final pressure of NO is x × 10^-8atm. The value of x is _______. [Use (10^6.13 = 1.34 x 10⁶)]
(Given: ${ΔG}_{f}^{o} {({NO}_{2})}_{(g)} = 52 KJ / mol, {ΔG}_{f}^{o} (NO)_{(g)} =$ 87 KJ/mol ) (report the value to nearest
integer)

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Detailed Solution

$\begin{array}{l} {NO}_{(g)} + \frac{1}{2} O_{2 (g)} ⇌ {NO}_{2} (g) \\ {ΔG}^{0} = {ΔG}_{f}^{0} {({NO}_{2})}_{(g)} - {ΔG}_{f}^{0} (NO)_{(g)} \end{array}$
= 52 - 87
$\begin{array}{l} {ΔG}^{0} = - 35 KJ \\ {ΔG}^{0} = - 2.303 RTlog K_{p} \end{array}$
-35 x 1000 = -2.303 x 8.314 x 298logK_p
$\Rightarrow \log K_{p} = 6.13 \Rightarrow K_{p} = 1.34 \times 10^{6}$
$\begin{array}{cccc} {NO}_{(g)} & + \frac{1}{2} O_{2 (g)} & ⇌ & {NO}_{2 (g)} \\ Initially & 1 & 11 / 4 & 0 \\ Equilibrium & p & 9 / 4 & 1 \end{array}$
$\begin{array}{l} K_{p} = 1.34 \times 10^{6} = \frac{P_{{NO}_{2}}}{P_{NO} \times P_{O_{2}}^{1 / 2}} \Rightarrow 1.34 \times 10^{6} = \frac{1}{p \times {(\frac{9}{4})}^{1 / 2}} \\ \Rightarrow p = 5 \times 10^{- 7} \Rightarrow p = 50 \times 10^{- 8} \end{array}$