Oxidation because oxygen causes the metal to corrode
Answer:
Kc = 0.075
Explanation:
The dissociation (α) is the initial quantity that ionized divided by the total dissolved. So, let's calling x the ionized quantity, and M the initial one:
α = x/M
x = M*α
x = 0.354M
For the stoichiometry of the reaction (2:1:1), the concentration of H₂ and I₂ must be half of the acid. So the equilibrium table must be:
2HI(g) ⇄ H₂(g) + I₂(g)
M 0 0 <em> Initial</em>
-0.354M +0.177M +0.177M <em>Reacts</em>
0.646M 0.177M 0.177M <em>Equilibrium</em>
The equilibrium constant Kc is the multiplication of the products' concentrations (elevated by their coefficients) divided by the multiplication of the reactants' concentrations (elevated by their coefficients):
![Kc = \frac{[H2]*[I2]}{[HI]^2}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BH2%5D%2A%5BI2%5D%7D%7B%5BHI%5D%5E2%7D)
Kc = 0.075
Answer:
The equilibrium pressure of NO2 is 0.084 atm
Explanation:
Step 1: Data given
A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3.
Kp = 0.0118
Step 2: The balanced equation
NO( g) + SO3( g) ⇌ NO2( g) + SO2( g)
Step 3: The initial pressures
p(NO) = 0.86 atm
p(SO3) = 0.86 atm
p(NO2) = 0 atm
p(SO2) = 0 atm
Step 4: The pressure at the equilibrium
For 1 mol NO we need 1 mol SO3 to produce 1 mol NO2 and 1 mol SO2
p(NO) = 0.86 -x atm
p(SO3) = 0.86 -xatm
p(NO2) = x atm
p(SO2) = x atm
Step 5: Define Kp
Kp = ((pNO2)*(pSO2)) / ((pNO)*(pSO3))
Kp = 0.0118 = x²/(0.86 - x)²
X = 0.08427
p(NO) = 0.86 -0.08427 = 0.77573 atm
p(SO3) = 0.86 -0.08427 = 0.77573 atm
p(NO2) = 0.08427 atm
p(SO2) = 0.08427 atm
The equilibrium pressure of NO2 is 0.08427 atm ≈ 0.084 atm
Answer:
Explanation:
Endothermic reactions absorb energy from the surrounding, but exothermic reactions release energy to the surrounding.
<span>The pH is given by the Henderson - Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = -log(</span><span>1.3 x 10^-5) + log(0.50/0.40)
pH = 4.98
The answer to this question is 4.98.
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