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inysia [295]
3 years ago
14

Consider the following reversible reaction. 2H2O(g)<—>2H2(g)+O2(g) What is the equilibrium constant expression for the giv

en system?
Chemistry
2 answers:
masha68 [24]3 years ago
7 0

The reaction: 2H2(g) + O2(g) → 2H2O(g), can be interpreted as: a. 2 moles of hydrogen gas reacts with 1 mole of oxygen gas to produce 2 moles of water.

docker41 [41]3 years ago
3 0

Answer : The equilibrium constant expression will be,

k_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}  

Explanation :

The given balanced chemical reaction is,

2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.  

The concentrations of pure solids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

So, the equilibrium constant expression will be,

k_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}

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3 years ago
Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t
Sedbober [7]

Answer:

Approximately 6.30\times 10^{-3}\;\rm mol.

Explanation:

The gallium here is likely to be produced from a \rm NaGaO_2\, (aq) solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral "\mathtt{(III)}" next to \rm Ga.  This numeral indicates that the oxidation state of the gallium in this solution is equal to +3. In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of \rm Ga\, (s).

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium \mathtt{(III)} solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s.

Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C.

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}.

It takes three moles of electrons to deposit one mole of gallium atoms \rm Ga\, (s). As a result, \rm 1.89\times 10^{-2}\; mol of electrons would deposit \displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol of gallium atoms \rm Ga\, (s).

8 0
3 years ago
The vapor pressure of pure water at 25 °c is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by
maxonik [38]

The vapour pressure of the solution is 23.4 torr.

Use <em>Raoult’s Law</em> to calculate the vapour pressure:  

<em>p</em>₁ = χ₁<em>p</em>₁°  

where  

χ₁ = the mole fraction of the solvent  

<em>p</em>₁ and <em>p</em>₁° are the vapour pressures of the solution and of the pure solvent  

The formula for vapour pressure lowering Δ<em>p</em> is  

Δ<em>p</em> = <em>p</em>₁° - <em>p</em>₁  

Δ<em>p</em> = <em>p</em>₁° - χ₁<em>p</em>₁° = p₁°(1 – χ₁) = χ₂<em>p</em>₁°  

where χ₂ is the mole fraction of the solute.  

<em>Step 1</em>. Calculate the <em>mole fraction of glucose </em>

<em>n</em>₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu  

<em>n</em>₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O  

χ₂ = <em>n</em>₂/(<em>n</em>₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62  

<em>Step 2</em>. Calculate the <em>vapour pressure lowering</em>  

Δ<em>p</em> = χ₂<em>p</em>₁° = 0.018 62 × 23.8 torr = 0.4430 torr  

<em>Step 3</em>. Calculate the <em>vapour pressure</em>  

<em>p₁</em> = <em>p</em>₁° - Δ<em>p</em> = 23.8 torr – 0.4430 torr = 23.4 torr

3 0
3 years ago
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