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andrew-mc [135]
3 years ago
11

At 258C, Kp 5 2.9 3 1023 for the reactionNH4OCONH21s2m 2NH31 g21 CO21 g2In an experiment carried out at 258C, a certain amount o

f NH4OCONH2 is placed in an evacuated rigid container and allowed to come to equilibrium. Calculate the total pressure in the container at equilibrium.
Chemistry
1 answer:
iragen [17]3 years ago
4 0

Answer:

0.27 atm

Explanation:

<em>At 25ºC, Kp = 2.9 x 10⁻³ for the reaction NH₄OCONH₂(s) ⇌ 2 NH₃(g) + CO₂(g). In an experiment carried out at 25ºC, a certain amount of NH₄OCONH₂ is placed in an evacuated rigid container and allowed to come to equilibrium. Calculate the total pressure in the container at equilibrium.</em>

Step 1: Make an ICE chart

Solid and liquids are ignored in ICE charts.

       NH₄OCONH₂(s) ⇌ 2 NH₃(g) + CO₂(g)

I                                            0             0

C                                        +2x           +x

E                                          2x             x

Step 2: Write the pressure equilibrium constant expression (Kp)

Kp = [NH₃]² × [CO₂]

Kp = (2x)² × x

2.9 × 10⁻³ = 4 x³

x = 0.090 atm

Step 3: Calculate the pressures at equilbrium

pNH₃ = 2x = 2(0.090 atm) = 0.18 atm

pCO₂ = x = 0.090 atm

The total pressure is:

P = 0.18 atm + 0.090 atm = 0.27 atm

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<h3>What is behavioral adaptation?</h3>

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2 years ago
Can anybody check my answer?
anzhelika [568]

Answer:

\boxed{\text{25. 20 L; 26. 49 K}}

Explanation:

25. Boyle's Law

The temperature and amount of gas are constant, so we can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Data:

\begin{array}{rcrrcl}p_{1}& =& \text{100 kPa}\qquad & V_{1} &= & \text{10.00 L} \\p_{2}& =& \text{50 kPa}\qquad & V_{2} &= & ?\\\end{array}

Calculations:

\begin{array}{rcl}100 \times 10.00 & =& 50V_{2}\\1000 & = & 50V_{2}\\V_{2} & = &\textbf{20 L}\\\end{array}\\\text{The new volume will be } \boxed{\textbf{20 L}}

26. Ideal Gas Law

We have p, V and n, so we can use the Ideal Gas Law to calculate the volume.

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p = 101.3 kPa

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n = 5 mol

R = 8.314 kPa·L·K⁻¹mol⁻¹

Calculation:

101.3 × 20 = 5 ×  8.314 × T

2026 = 41.57T

T = \dfrac{2026}{41.57} = \textbf{49 K}\\\\\text{The Kelvin temperature is }\boxed{\textbf{49 K}}

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