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sertanlavr [38]
3 years ago
10

Consider the following reaction where Kc = 6.50×10-3 at 298 K: 2NOBr(g) 2NO(g) + Br2(g) A reaction mixture was found to contain

9.83×10-2 moles of NOBr(g), 5.44×10-2 moles of NO(g), and 4.13×10-2 moles of Br2(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc
Chemistry
1 answer:
Aleks [24]3 years ago
8 0

Answer:

This reaction isn't yet at an equilibrium. It must shift in the direction of the reactant (namely \rm NOBr\; (g)) in order to reach an equilibrium.

For this mixture, the reaction quotient is Q_c = 0.0126.

Explanation:

A reversible reaction is at equilibrium if and only if its reaction quotient Q_c is equal to the equilibrium constant K_c.

Start by calculating the equilibrium quotient Q_c of this reaction. Given the reaction:

\rm 2\; NOBr\; (g) \rightleftharpoons 2\; NO\; (g) + Br_2\; (g).

Let [\mathrm{NOBr\; (g)}], [\mathrm{NO\; (g)}], and [\mathrm{Br_2\; (g)}] denote the concentration of the three species. The formula for the reaction quotient of this system will be:

\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2}.

(Note, that in this formula, both [\mathrm{NO\; (g)}] and [\mathrm{NOBr\; (g)}] are raised to a power of two. That corresponds to the coefficients in the balanced reaction.)

Calculate the reaction quotient given the concentration of each species:

\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2} \approx 1.26\times 10^{-2} = 0.0126.

(Note that the unit is ignored.)

Apparently, Q_c > K_c. Since Q_c and K_c are not equal, this reaction is not at an equilibrium. If external factors like temperature stays the same,

Keep in mind that Q_c denotes a quotient. To reduce the value of a quotient, one may:

  • reduce the value of the numerator,
  • increase the value of the denominator, or
  • both.

In Q_c, that means reducing the concentration of the products while increasing the concentration of the reactants. In other words, the system needs to shift in the direction of the reactants before it could reach an equilibrium.

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Answer:

Explanation:

energy for photon is calculated in same wasy as for electromagnetic radiation

energy for electromagnetic radiation = hf

where f is the frequency of photon

h is Planck's constant = h = 4.14 × 10−15 eV · s.

thus

energy of photon = 4.14 × 10−15 eV · s *  2.36 x 10-19 Hz

energy of photon = 9.77 * 10−(-15+ -19) eV

energy of photon = 9.77 * 10−34eV        answer

6 0
2 years ago
A sample of a compound that contains only the elements C, H, and N is completely burned in O₂ to produce 44.0 g of CO₂, 45.0 g o
koban [17]

Answer:

CH₅N

Explanation:

In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:

(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C

Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:

(45.0 g)(mol/18.02g) = 2.497...mol H₂O

Moles of H is found using the molar ratio of 2H:1H₂O:

(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H

The ratio of H to C in the compound is:

(4.994...mol H)/(0.99977... mol C) = 5 H:C

Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.

3 0
2 years ago
If 1.02 g of nickel reacted with 750. mL of 0.112 M hydrobromic acid, how much of each will be present at the end of the reactio
kati45 [8]

Answer:

35.1% is percent yield

Explanation:

<em>Full question: Assume no volume change.  If you formed 0.0910 atm of gas, what is the percent yield?</em>

<em />

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

<em>Moles Ni (Molar mass: 58.69g/mol):</em>

1.02g * (1mol / 58.69g) = 0.01738moles Ni

<em>Moles HBr:</em>

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

<em>Where P is pressure (0.0910atm)</em>

<em>V is volume (2.50L)</em>

<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)</em>

<em>And n are moles</em>

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

<em />

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

<h3>35.1% is percent yield</h3>
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