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mylen [45]
3 years ago
5

NH4CO2NH2(s) equilibrium reaction arrow 2 NH3(g) + CO2(g) Ammonium carbamate decomposes according to the equation above. At 40°C

, an equilibrium mixture for this reaction contains 0.0640 atm CO2 and 0.370 atm NH3. What is the value of the equilibrium constant Kp for the reaction?
Chemistry
1 answer:
V125BC [204]3 years ago
4 0

Answer:

Kp = 8.76×10⁻³

Explanation:

We determine the carbamate decomposition in equilibrium:

NH₄CO₂NH₂ (s) ⇄ 2NH₃(g) + CO₂(g)

Let's build the expression for Kp

Kp = (Partial pressure NH₃)² . Partial pressure CO₂

We do not consider, the carbamate because it is solid and we only need the partial pressure from gases

Kp = (0.370atm)² . 0.0640 atm

Kp = 8.76×10⁻³

Remember Kp does not carry units

 

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Answer:

V_2=18 \ L \ H_2

General Formulas and Concepts:

<u>Chemistry - Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
  • Charles' Law: \frac{V_1}{T_1} =\frac{V_2}{T_2}

Explanation:

<u>Step 1: Define</u>

Initial Volume: 5.0 L H₂ gas

Initial Temp: 273 K

Final Temp: 985 K

Final Volume: ?

<u>Step 2: Solve for new volume</u>

  1. Substitute:                    \frac{5 \ L \ H_2}{273 \ K} =\frac{x \ L \ H_2}{985 \ K}
  2. Cross-multiply:             (5 \ L \ H_2)(985 \ K) = (x \ L \ H_2)(273 \ K)
  3. Multiply:                        4925 \ L \ H_2 \cdot K = 273x \ L \ H_2 \cdot K
  4. Isolate <em>x</em>:                       18.0403 \ L \ H_2 = x
  5. Rewrite:                         x=18.0403 \ L \ H_2

<u>Step 3: Check</u>

<em>We are given 2 sig figs as the smallest. Follow sig fig rules and round.</em>

<em />18.0403 \ L \ H_2 \approx 18 \ L \ H_2<em />

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