Answer:
The total pressure at equilibrium is 0.07503 atm
Explanation:
The partial pressure of the product at equilibrium will be calculated as follows;
Kp = Kc[RT]³
given;
equilibrium constant Kc = 1.58 X 10⁻⁸
gas constant R = 0.0821 L.atm/mol.K
temperature T = (250 +273) = 523 k
Kp = 1.58 X 10⁻⁸ *(0.0821)³ *(523)³ = 1.251 X 10⁻³
NH₂COONH₄(s) ⇌ 2NH₃(g) + CO₂
NH₂COONH₄(s): Kp = 0, since it is in solid state
2NH₃(g) + CO₂: Kp = 1.251 X 10⁻³
I.C.E Analysis on the product
2NH₃(g) CO₂
I : 0 0
C : 2x x
E : (2x-0) (x-0)
At equilibrium, E: (2x-0)(x-0) = 1.251 X 10⁻³
(2x)(x) = 1.251 X 10⁻³
2x² = 1.251 X 10⁻³
x² = (1.251 X 10⁻³)/2
x² = 6.255 X 10⁻⁴
x = √(6.255 X 10⁻⁴)
x = 0.02501 atm
Partial pressure of 2NH₃(g) = 2x = 2(0.02501 atm) = 0.05002 atm
Partial pressure of CO₂ = x = 0.02501 atm
Total pressure = P(NH₃(g)) +P(CO₂)
Total pressure = 0.05002 atm + 0.02501 atm = 0.07503 atm
Therefore, the total pressure at equilibrium is 0.07503 atm
From data provided, the total pressure at equilibrium is 0.203 atm.
The total pressure, Ptotal at equilibrium is calculated from the equation of the reaction given below:
From the equation of the reaction, If x moles of ammonium carbamate decomposes, it will produce 2x moles of NH3(g) and x moles of CO2(g).
Ammonium carbamate is a solid, and so it does not appear in the expression for Kc.
Kc = 1.58 × 10^-8
Therefore:
Kc = [NH3(g)]^2[CO2(g)]
Kc = (2x)^2(x) = 4x^3
1.58 × 10-8 = 4x^3
Thus, x = 0.00158 M
Hence:
[NH3(g)] = 2 × 0.00158 = 0.00316 M
[CO2(g)] = 0.00158 M
From the ideal gas equation:
where
Also, concentration is given by:
Therefore, P = cRT.
Substituting and calculating for Ptotal:
Ptotal = (0.00316 M) × RT + (0.00158 M) × RT
= ((0.00316 + 0.00158) M) × (0.08206) × ((523) K)
Ptotal = 0.203 atm
Therefore, the total pressure at equilibrium is 0.203 atm.
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