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ra1l [238]
3 years ago
11

At equilibrium, both the forward reaction and the reverse reaction occur simultaneously and at the same rate. Since the rates ar

e the same, the concentrations of the reactants and products do not change. An equilibrium constant expression can be developed, which is a mathematical expression indicating the relationship between the reactants and the products. For reactions with gases, the value of Kp can be determined using the pressures of each substance to the appropriate power. The values of Kc and Kp are related by the equation, Kp = Kc(RT)n. At 298 K, Kc = 18.3 for the following reaction. What is the value of K(p)?
C₄(s) + 4 O₂(g) ⇌ 4 CO₂(g)
Chemistry
1 answer:
Alja [10]3 years ago
6 0

Answer:

The value of the gas equilibrium constant Kp is 18.3

Explanation:

Step 1: Data given

Kp= Kc(RT)^Δn

Temperature = 298 K

Kc = 18.3

Step 2: The balanced equation

C₄(s) + 4 O₂(g) ⇌ 4 CO₂(g)

Step 3: Calculate Kp

Kp= Kc(RT)^Δn

⇒with Kp = the gas equilibrium constant =  TO BE DETERMINED

⇒with Kc = the equilibrium constant = 18.3

⇒R = the gas constant = 0.08206 L* atm/mol*K

⇒T = the temperature = 298 K

⇒Δn = mol gas phase product – mol gas phase reactant = 4-4 =0

Kp = 18.3 * (0.08206*298)^0

Kp = 18.3 * 1

Kp = 18.3

The value of the gas equilibrium constant Kp is 18.3

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If 24.3 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over?2 NO (g) + O₂ (g) → 2
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Explanation:

2 NO (g) + O₂ (g) ----> 2 NO₂ (g)

24.3 g of NO are reacting with 13.8 g of O₂. First we can convert the mass of theses samples into moles using their molar masses.

molar mass of O = 16.00 g/mol

molar mass of N = 14.01 g/mol

molar mass of NO = 16.00 g/mol + 14.01 g/mol

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molar mass of O₂ = 2 * 16.00 g/mol

molar mass of O₂ = 32.00 g/mol

moles of NO = 24.3 g * 1 mol/(30.01 g)

moles of NO = 0.810 moles

moles of O₂ = 13.8 g * 1 mol/(32.00 g)

moles of O₂ = 0.431 moles

Now, to determine the limiting reactant or the excess reactant we can find the number of moles of O₂ that will react with 0.810 moles of NO and the number of moles of NO that will react with 0.431 moles of O₂.

According to the coefficients of the reaction 2 moles of NO will react with 1 mol of O₂. Let's use that relationship to find the limiting reagent.

2 moles of NO = 1 mol of O₂

moles of O₂ = 0.810 moles of NO * 1 mol of O₂/(2 moles of NO)

moles of O₂ = 0.405 moles

moles of NO = 0.431 moles of O₂ * 2 moles of NO/(1 mol of O₂)

moles of NO = 0.862 moles

We found that we need 0.405 moles of O₂ to completely react with 0.810 moles of NO. Or, we need 0.862 moles of NO to completely react with ours 0.431 moles of NO.

We can say that NO is limiting our reaction and O₂ is in excess.

Only 0.405 moles of O₂ will react with 0.810 moles of NO. But we had 0.431 moles of it. Let's find the excess.

Excess of O₂ = 0.431 moles - 0.405 moles

Excess of O₂ = 0.026 moles

Answer: 0.026 moles is the number of moles of oxygen that will be left over.

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