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Sati [7]
3 years ago
12

At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o

f HBr, H2, and Br2 are 1.0 x 10^-2 atm, 2.0 x 10^-4 atm, and 2.0 x 10^-4 atm, respecivtely, what is the equilbrium partial pressure of H2?
Chemistry
1 answer:
Damm [24]3 years ago
4 0

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

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3 years ago
Cesium has a radius of 272 pm and crystallizes in a body-centered cubic structure. What is the edge length of the unit cell
olchik [2.2K]

Answer: Edge length of the unit cell = 628pm

Explanation: For a body centred cubic structured system, the relationship between the edge length of the unit cell and radius of the atoms in the structure is

Edge length of Unit cell (a) = (4R)/(√3)

R = 272pm = (272 × (10^-12))m = (2.72 × (10^-10))m

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4 0
2 years ago
What is the molar mass of CaCL2•2 H2O
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Answer:

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3 years ago
Biacetyl, the flavoring that makes margarine taste "just like butter," is extremely stable at room temperature, but at 200°C it
Sholpan [36]

Answer:

3.91 minutes

Explanation:

Given that:

Biacetyl breakdown with a half life of 9.0 min after undergoing first-order reaction;

As we known that the half-life for first order is:

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where;

k = constant

The formula can be re-written as:

k = \frac{0.693}{t__{1/2}}

k = \frac{0.693}{9.0 min}

k = 0.077 min^{-1}

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Also, the amount of butter flavor retained at 200°C (N_t)= 74%

The rate constant k = 0.077 min^{-1}

To determine how long can the food be heated at this temperature and retain 74% of its buttery flavor; we use the formula:

\frac{N_t}{N_0}= -kt

t = - (\frac{1}{k}*In\frac{N_t}{N_0}  )

Substituting our values; we have:

t = - (\frac{1}{0.077}*In\frac{74}{100}  )

t = 3.91 minutes

∵ The time needed for the food to be heated at this temperature and retain 74% of its buttery flavor is 3.91 minutes

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