Respuesta :
Answer:
Individual mole fractions of all the species of the all reaction is as follows.
(a)
[tex]y_{NH_{3}}=\frac{2+(-4)\epsilon}{7+\epsilon}[/tex]
[tex]y_{O_{2}}=\frac{5+(-5)\epsilon}{7+\epsilon}[/tex]
[tex]y_{NO}=\frac{0+(4)\epsilon}{7+\epsilon}=\frac{4\epsilon}{7+\epsilon}[/tex]
[tex]y_{H_{2}O}=\frac{0+6\epsilon}{7+\epsilon}[/tex]
(b)
[tex]y_{H_{2}S}=\frac{3+(-2)\epsilon}{8- \epsilon}[/tex]
[tex]y_{O_{2}}=\frac{5+(-3)\epsilon}{8- \epsilon}[/tex]
[tex]y_{H_{2}O}=\frac{2\epsilon}{8- \epsilon}[/tex]
[tex]y_{SO_{2}}=\frac{2\epsilon}{8- \epsilon}[/tex]
(c)
[tex]y_{NO_{2}}=\frac{3+(-6)\epsilon}{8+5\epsilon}[/tex]
[tex]y_{NH_{3}}=\frac{4+(-8)\epsilon}{8+5\epsilon}[/tex]
[tex]y_{N-{2}}=\frac{1+7\epsilon}{8+5 \epsilon}[/tex]
[tex]y_{H_{2}O}=\frac{12\epsilon}{8+5\epsilon}[/tex]
Explanation:
(a)
Initial number of moles of [tex]NH_{3}[/tex] and [tex]O_{2}[/tex] are 2 mol and 5 mol respectively.
The given chemical reaction is as follows.
[tex]4NH_{3}(g)+5O_{2}(g)\rightarrow 4NO(g)+6H_{2}O[/tex]
The stoichiometric numbers are as follows.
[tex]v_{NH_{3}}=-4[/tex]
[tex]v_{O_{2}}=-5[/tex]
[tex]v_{NO}=4[/tex]
[tex]v_{H_{2}O}=6[/tex]
The total number of moles initially present -7
[tex]\Sigma v_{i}\epsilon = (-4-5+4+6)= 1\epsilon[/tex]
The expression for the mole fraction of species"i" is as follows.
[tex]y_{i}=\frac{(n_{i_{o}})+(v_{i}\epsilon) }{n_{o}+v\epsilon}[/tex]
The individual mole fractions of all the species are as follows.
[tex]y_{NH_{3}}=\frac{2+(-4)\epsilon}{7+\epsilon}[/tex]
[tex]y_{O_{2}}=\frac{5+(-5)\epsilon}{7+\epsilon}[/tex]
[tex]y_{NO}=\frac{0+(4)\epsilon}{7+\epsilon}=\frac{4\epsilon}{7+\epsilon}[/tex]
[tex]y_{H_{2}O}=\frac{0+6\epsilon}{7+\epsilon}[/tex]
(b)
Initial number of moles of [tex]H_{2}S[/tex] and [tex]O_{2}[/tex] are 3 mol and 5 mol respectively.
The given chemical reaction is as follows.
[tex]2H_{2}S(g)+3O_{2}(g)\rightarrow 2H_{2}O(g)+2SO_{2}[/tex]
The stoichiometric numbers are as follows.
[tex]v_{H_{2}S}=-2[/tex]
[tex]v_{O_{2}}=-3[/tex]
[tex]v_{H_{2}O}=2[/tex]
[tex]v_{SO_{2}=2[/tex]
The total number of moles initially present -8
[tex]\Sigma v_{i}\epsilon = (-2-3+2+2)= - \epsilon[/tex]
The expression for the mole fraction of species"i" is as follows.
[tex]y_{i}=\frac{(n_{i_{o}})+(v_{i}\epsilon) }{n_{o}+v\epsilon}[/tex]
The individual mole fractions of all the species are as follows.
[tex]y_{H_{2}S}=\frac{3+(-2)\epsilon}{8- \epsilon}[/tex]
[tex]y_{O_{2}}=\frac{5+(-3)\epsilon}{8- \epsilon}[/tex]
[tex]y_{H_{2}O}=\frac{2\epsilon}{8- \epsilon}[/tex]
[tex]y_{SO_{2}}=\frac{2\epsilon}{8- \epsilon}[/tex]
(c)
Initial number of moles of [tex]NO_{2}[/tex], [tex]NH_{3}[/tex]and [tex]N_{2}[/tex] are 3 mol,4 mol and 1 mol respectively.
The given chemical reaction is as follows.
[tex]6NO_{2}(g)+8NH_{3}(g)\rightarrow 7N_{2}(g)+12H_{2}O[/tex]
The stoichiometric numbers are as follows.
[tex]v_{NO_{2}}=-6[/tex]
[tex]v_{NH_{3}}=-8[/tex]
[tex]v_{N_{2}}=7[/tex]
[tex]v_{H_{2}O}=12[/tex]
The total number of moles initially present -8
[tex]\Sigma v_{i}\epsilon = (-6-8+7+12)= 5 \epsilon[/tex]
The expression for the mole fraction of species"i" is as follows.
[tex]y_{i}=\frac{(n_{i_{o}})+(v_{i}\epsilon) }{n_{o}+v\epsilon}[/tex]
The individual mole fractions of all the species are as follows.
[tex]y_{NO_{2}}=\frac{3+(-6)\epsilon}{8+5\epsilon}[/tex]
[tex]y_{NH_{3}}=\frac{4+(-8)\epsilon}{8+5\epsilon}[/tex]
[tex]y_{N-{2}}=\frac{1+7\epsilon}{8+5 \epsilon}[/tex]
[tex]y_{H_{2}O}=\frac{12\epsilon}{8+5\epsilon}[/tex]
