Answer:
Explanation:
The balanced equation is
I₂(g) + Br₂(g) ⇌ 2IBr(g)
Data:
Kc = 8.50 × 10⁻³
n(IBr) = 0.0600 mol
V = 1.0 L
1. Calculate [IBr]
![\text{[IBr]} = \dfrac{\text{0.0600 mol}}{\text{1.0 L}} = \text{0.0600 mol/L}](https://tex.z-dn.net/?f=%5Ctext%7B%5BIBr%5D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.0600%20mol%7D%7D%7B%5Ctext%7B1.0%20L%7D%7D%20%3D%20%5Ctext%7B0.0600%20mol%2FL%7D)
2. Set up an ICE table.
3. Calculate [I₂]
![\begin{array}{rcl}K_{\text{c}}&=&\dfrac{\text{[IBr]}^{2}} {\text{[I$_{2}$][Br]$_{2}$}}\\\\8.50 \times 10^{-2}&=&{\dfrac{(0.0600 - 2x)^{2}}{x^{2}}}& &\\\\0.2915x & = &{\dfrac{0.0600 - 2x}{x}}& &\\\\0.2915x & = &0.0600 - 2x\\\\2.2915x & = & 0.0600\\x & = & \textbf{0.026 18 mol/L}\\\end{array}\\](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7DK_%7B%5Ctext%7Bc%7D%7D%26%3D%26%5Cdfrac%7B%5Ctext%7B%5BIBr%5D%7D%5E%7B2%7D%7D%20%7B%5Ctext%7B%5BI%24_%7B2%7D%24%5D%5BBr%5D%24_%7B2%7D%24%7D%7D%5C%5C%5C%5C8.50%20%5Ctimes%2010%5E%7B-2%7D%26%3D%26%7B%5Cdfrac%7B%280.0600%20-%202x%29%5E%7B2%7D%7D%7Bx%5E%7B2%7D%7D%7D%26%20%26%5C%5C%5C%5C0.2915x%20%26%20%3D%20%26%7B%5Cdfrac%7B0.0600%20-%202x%7D%7Bx%7D%7D%26%20%26%5C%5C%5C%5C0.2915x%20%26%20%3D%20%260.0600%20-%202x%5C%5C%5C%5C2.2915x%20%26%20%3D%20%26%200.0600%5C%5Cx%20%26%20%3D%20%26%20%5Ctextbf%7B0.026%2018%20mol%2FL%7D%5C%5C%5Cend%7Barray%7D%5C%5C)
4. Convert the temperature to kelvins
T = (150 + 273.15) K = 423.15 K
5. Calculate p(I₂)
Ch4 is the lowest boiling point
Answer:
V = 10.3 L
Explanation:
Given data:
Mass of methane = 6.40 g
Volume of CO₂ produced = ?
Temperature = 35°C (35+273 = 308 K)
Pressure = 100.0 KPa (100.0/101 = 0.98 atm)
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of CH₄:
Number of moles = mass/molar mass
Number of moles = 6.40 g/ 16 g/mol
Number of moles = 0.4 mol
Now we will compare the moles of CO₂ with CH₄.
CH₄ : CO₂
1 : 1
0.4 : 0.4
Volume of CO₂:
Formula:
PV = nRT
0.98 atm ×V = 0.4 mol ×0.0821 atm.L/mol.K × 308 K
0.98 atm ×V = 10.11 atm.L
V = 10.11 atm.L /0.98 atm
V = 10.3 L
It is tasteless and colorless.
