Giúp em ạ Huhuhuhuu

Đáp án:

$N_2O_4(k)\rightleftharpoons 2NO_2(k)$
a,

$n_{N_2O_4\text{phân li}}=a\alpha$ 

$\Rightarrow n_{NO_2}=2a\alpha$ 

$\Rightarrow n_{N_2O_4}= a-a\alpha=a(1-\alpha)$ 

$n_{\text{khí tổng}}=2a\alpha+a-a\alpha=a+a\alpha=a(1+\alpha)$ 

b, 

$P_{NO_2}=P_{\text{hệ}}.X_{NO_2}=P.\dfrac{2a\alpha}{a(1+\alpha)}=\dfrac{2P\alpha}{1+\alpha}$ 

$P_{N_2O_4}=P_{\text{hệ}}.X_{N_2O_4}=P.\dfrac{a(1-\alpha)}{a(1+\alpha)}=\dfrac{P(1-\alpha)}{1+\alpha}$ 

$K_P=\dfrac{P_{NO_2}^2}{P_{N_2O_4}}=\dfrac{ 4P^2\alpha^2}{(1+\alpha)^2}.\dfrac{1+\alpha}{P(1-\alpha)}=\dfrac{4P\alpha^2}{1-\alpha^2}$ 

c, 

$n_{N_2O_4\text{bđ}}=\dfrac{1,588}{92}=0,0173(mol)$ 

$n_{\text{khí tổng}}=\dfrac{0,5.1}{0,082.298}=0,0205(mol)$ 

$\Rightarrow \alpha=\dfrac{n_{\text{khí tổng}}}{a}-1= 0,185=18,5\%$ 

$P_{NO_2}=\dfrac{2.1.0,185}{1+0,185}=0,3122atm$
$P_{N_2O_4}=P-P_{NO_2}=0,6878atm$