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Fed [463]
4 years ago
8

Consider the following reaction where Kc = 34.5 at 1150 K:2SO2(g) + O2(g) 2SO3(g)A reaction mixture was found to contain 4.77×10

-2 moles of SO2(g), 4.55×10-2 moles of O2(g) and 4.37×10-2 moles of SO3(g), in a 1.00 Liter container.Indicate True (T) or False (F) for each of the following:1. In order to reach equilibrium SO3(g) must be produced.2. In order to reach equilibrium Kc must increase.3. In order to reach equilibrium SO2 must be produced.4. Qc is greater than Kc.5. The reaction is at equilibrium. No further reaction will occur.
Chemistry
1 answer:
sukhopar [10]4 years ago
3 0

Answer:

Explanation:

[ so₃] = 4.37 x 10⁻²

[so₂] = 4.77 x 10⁻²

[ o₂] = 4.55 x 10⁻²

Qc = (4.37)²x10⁻⁴ /(4.77)².(4.55) x 10⁻⁶ =18.44

Qc is less than Kc hence in order to reach equilibrium more of so₃ will be produced . Statement 1 is true.

Kc is always constant . Statement 2 is false.

Statement 3 is false because statement 1 is true.

Qc Is smaller than Kc . So statement 4 is false.

The reaction is not in equilibrium. Statement 5 is false.

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A hypothesis is an experimental question that can later be proven correct or incorrect by scientific evidence.

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Elements in the periodic are listed according to
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4 years ago
When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit
pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq).

How many moles of this process?

Relative atomic mass from a modern periodic table:

  • K: 39.098;
  • N: 14.007;
  • O: 15.999.

Molar mass of \text{KNO}_3:

M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of the process = Number of moles of \text{KNO}_3 dissolved:

\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}.

What's the enthalpy change of this process?

Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ} for 0.212162\;\text{mol}. By convention, the enthalpy change \Delta H measures the energy change for each mole of a process.

\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}.

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

4 0
3 years ago
Why is there a difference in the conductivity of pure solid NaC1 and of the 1.0 M NaC1 soulution
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7 0
4 years ago
4.33 g of 3-hexanol were obtained from 5.84 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles o
Vilka [71]

<u>Answer:</u> The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol

The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:

\text{hex-3-ene}+H_2O\xrightarrow []{10\% H_2SO_4} \text{3-hexanol}

By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

So, 0.0711 moles of hex-3-ene will produce = \frac{1}{1}\times 0.0711=0.0711mol of 3-hexanol

Now, calculating the mass of 3-hexanol from equation 1, we get:

Molar mass of 3-hexanol = 102.2 g/mol

Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g

To calculate the percentage yield of 3-hexanol, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%

Hence, the amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

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3 years ago
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