Answer:
[tex]k_c =[/tex] 1. 1 × 10⁻²
Explanation:
Given that:
Temperature = 335 ° C = (335+ 273)K = 608
Pressure = 0.750 atm
Volume = 1 Litre
number of moles of NO2 = ???
Rate Constant =0.0821 L atm /K/mol
Using the Ideal gas equation
PV = nRT
n = [tex]\frac{PV}{RT}[/tex]
n = [tex]\frac{0.75*1}{0.0821*608}[/tex]
n = 0.015
n = 1.5 × 10⁻² mole
Density = 0.525 g/L
The equation for the reaction can be illustrated as:
2NO2(g) ⇌ 2NO(g) + O2(g)
For the ICE table; we have:
Initial x 0 0
Change -2y + 2y +y
Equilibrium (x - 2y) 2y y
Total moles at equilibrium = (x-2y)+2y+y
= x + y moles
However,
1.5 × 10⁻² mole of the mixture has a mass of 0.525 g
i.e x + y moles = 1.5 × 10⁻² mole
Now, molar mass of 1 mole of NO2 = 46g/mol
Since number of moles = [tex]\frac{mass}{molar mass}[/tex]
mass of (x-2y) moles = 46 × (x-2y) g
Molar mass of NO = 30 g/mol
Also, mass of NO = 2y × 30 = 60y
Molar mass of O2 = 32 g/mol
Mass of O2 = y × 32 = 32y
Total mass = ( 46x - 90y)+60y+32y = 0.525
46x = 0.525
x = [tex]\frac{0.525}{46}[/tex]
x = 0.0114
x = 1.14 × 10⁻²
x + y moles = 1.5 × 10⁻²
y = 1.5 × 10⁻² - 1.14 × 10⁻²
y = 0.0036
y = 3.6 × 10⁻³
At equilibrium
[NO2] = ( 1.14 - 2(0.36))× 10⁻² = 4.2 × 10⁻³ M
[NO] = 2 ( 3.6 × 10⁻³) = 7.2 × 10⁻³ M
[O2] = 3.6 × 10⁻³ M
[tex]k_c = \frac{[NO]^2[O_2]}{[NO_2]^2}[/tex]
[tex]k_c = \frac{(7.2*10^{-3})^2(3.6*10^{-3})}{(4.2*10^-3)^2}[/tex]
[tex]k_c =[/tex] 0.011
[tex]k_c =[/tex] 1. 1 × 10⁻²
Answer:
0.97.
Explanation:
So, let us first write out the balanced equilibrium equation below;
2 NO2(g) <============> 2 NO(g) + O2(g).
So, from the above equation of reaction 2 moles of NO2 gives us 3 moles of NO and one mole of O2.
Let us say that initially the concentration NO2 = 1 and that that NO and O2 is zero respectively. Then, at specific time, t = t the concentration of NO2 = 1 - 2x and the concentration of NO and O2 = 2x and x respectively.
So, kc = [2x]^2 [x] / [ 1 - 2x]^2.
Now, we have two unknown variables that is the value of x and kc. So, we will have to find the value of x first before we can proceed to solve for kc.
Recall the ideal gas law equation which is; PV = nRT. Where P= pressure, V= volume, R= gas constant, n= number of moles and T = temperature.
Also, n= mass/ molar mass.
So, PV = (mass/ molar mass) × R × T.
We are given the density of the gas mixture to be = 0.525 g/L. Where density = mass/ volume. And pressure = 0.750 atm.
Hence, P =( density × R × T)/ Molar mass.
Molar mass = 0.525 × 0.0821 × (335° C + 273)/ 0.750.
Molar mass= 34.9 g/ mol.
The next thing to do is to find the value of x.
Therefore;
Molar mass of the mixture= Molar mass of NO2 × (1 -2x / 1 + x) + molar mass of NO ( 2x/ 1 + x) + molar mass of O2 ( x/ 1 + x).
Therefore, solving for x gives;
34.9 = 46/ 1 + x.
34.9 + 34.9 x = 46.
34.9 x = 46 - 34.9.
x= 0.318.
So, kc = (0.6361)^2 (0.318) / (0.3639)^2.
Kc= 0.4046 × 0.318/ 0.1324.
kc = 0.97.
