# Question 5ecc9

Jun 3, 2017

${K}_{c} = 1.96 x {10}^{5}$

#### Explanation:

$2 {N}_{2} {O}_{5} \left(s\right) r i g h t \le f t h a r p \infty n s 4 N {O}_{2} \left(g\right) + {O}_{2} \left(g\right)$

Use DeltaG^o = -RTlnK_c" => ${K}_{c} = {e}^{- \left(\Delta {G}^{o} \text{/"RT}\right)}$

$\Delta {G}^{o} = \Delta {H}^{o} - T \Delta {S}^{o}$

$T = 298.15 K$

$\Delta {H}^{o} = \Sigma n \cdot \Delta {H}_{f}^{o} \left(P r o \mathrm{du} c t s\right) - \Sigma n \cdot \Delta {H}_{f}^{o} \left(R e a c \tan t s\right)$
$= \left[4 \Delta {H}_{f}^{o} \left(N {O}_{2} \left(g\right)\right) + 1 \Delta {H}_{f}^{o} \left({O}_{2} \left(g\right)\right)\right] - \left[2 \Delta {H}_{f}^{o} \left({N}_{2} {O}_{5} \left(s\right)\right)\right]$
$= \left[4 \left(33.18\right) + 1 \left(0\right)\right] K j - \left[2 \left(11.3\right)\right] K j$
$= + 110.12 K j$

$\Delta {S}^{o} = \Sigma n \cdot {S}^{o} \left(P r o \mathrm{du} c t s\right) - \Sigma n \cdot {S}^{o} \left(R e a c \tan t s\right)$
$= \left[4 {S}^{o} \left(N {O}_{2} \left(g\right)\right) + 1 {S}^{o} \left({O}_{2} \left(g\right)\right)\right] - \left[2 {S}^{o} \left({N}_{2} {O}_{5} \left(s\right)\right)\right]$
$= \left[4 \left(239.95\right) + 1 \left(205.03\right)\right] \frac{J}{K} - \left[2 \left(347.19\right)\right] \frac{J}{K}$
$= + 470.45 \frac{J}{K} = + 0.47045 \frac{\text{Kj}}{K}$

$\Delta {G}^{o} a t \left(298.15 K\right) :$
$\Delta {G}^{o} = \Delta {H}^{o} - T \Delta {S}^{o}$
$= \left[110.12 K j\right] - \left(298.15 K\right) \left[0.47045 \frac{\text{Kj}}{K}\right]$
$= - 30.14 K j$

DeltaG^o = -RTlnK_c"
=> $\ln {K}_{c} = - \left(\Delta {G}^{o} \text{/"R*T}\right)$$= - \left[\frac{- 30.14}{0.008314 \cdot 298.15}\right] \text{mole}$
= 12.18322mol"#; drop units in final value of ${K}_{c}$

${K}_{c} = {e}^{12.18322}$ =$1.96 x {10}^{5}$