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Veronika [31]
4 years ago
8

Which law states that the volume of a gas is proportional to the moles of the gas when pressure and temperature are kept constan

t
Chemistry
1 answer:
barxatty [35]4 years ago
3 0

Answer:

  • <em>The law that states that the volume of a gas is proportional to the moles of the gas when pressure and temperature are kept constant is </em><u>Avogadro's law.</u>

Explanation:

Pressure, temperature, volume, and amount of gases are related.

There are four basic gas laws: Boyle's law, Charles' law, Gay-Lussac's law and Avogadro's law.

The combination of the first three above mentioned laws gives rise to the combined law of gases. When Avogadro's law is incorporated, ideal gas equation is obtained.

These are the laws:

<u>1) Boyle's law: </u>at constant temperature, the pressure are volume of the a fiexed amount of gases are inversely related:

  • PV = constant
  • P₁V₁ = P₂V₂

<u>2) Charle's law:</u> at constant pressure, the volume and temperarure of a fixed amount of gases are directly (proportionaly) related:

  • P/T = constant
  • P₁ / T₁ = P₂ / T₂

<u>3) Gay-Lussac's law</u>: if the volume of a fixed amount of gas is kept constante, the pressure and temperature are directly related:

  • P / T = constant
  • P₁ / T₁ = P₂ / T₂

<u>4) Combined law</u>: shows the relation between pressure, volume, and temperature for a fixed amount of gas:

  • PV / T = constant
  • P₁V₁ / T₁ = P₂V₂ / T₂

In all previous law the numerical value of the temperature must be expressed in absolute scale: kelvin.

<u>5) Avogadro's law</u>: equal volume of gases at the same temperature and pressure contain the same number of particles (atoms or molecules).

Using n for the number of moles of gas particles:

  • V/n = constant
  • V₁ / n₁ = V₂ / T₂
  • 1 mol of gas at T = 0°C (273.15 K) and P = 1 atm occupies aproximately 22.4 liter.

<u>6) Ideal gas law</u>:

  • PV / nT = constant
  • R = Universal constant of gases
  • PV / nT = R
  • PV = nRT
  • T expressed in absolute scale (Kelvin)
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There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
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To omit acetylene from the net chemical reaction, we multiply Equation (1) by 6.

<u>Equation 1:</u>  6CaC_2(s)+12H_2O(g)\rightarrow 6C_2H_2(g)+6Ca(OH)_2(s)

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<u>Net chemical equation:</u>   6CaC_2(s)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)+6Ca(OH)_2(s)

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6 0
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The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T
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Answer :

(a) The rate law for the reaction is:

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(b) The value of rate constant is, 60.4M^{-1}s^{-1}

(c) rate of the reaction is 6.52\times 10^{-5}Ms^{-1}

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

OCl^-+I^-\rightarrow OI^-+Cl^-

Rate law expression for the reaction:

\text{Rate}=k[OCl^-]^a[I^-]^b

where,

a = order with respect to OCl^-

b = order with respect to I^-

Expression for rate law for first observation:

1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b ....(1)

Expression for rate law for second observation:

2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b ....(2)

Expression for rate law for third observation:

2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b ....(3)

Dividing 1 from 2, we get:

\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1

Dividing 1 from 3, we get:

\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1

Thus, the rate law becomes:

\text{Rate}=k[OCl^-]^a[I^-]^b

a  = 1 and b = 1

\text{Rate}=k[OCl^-]^1[I^-]^1

Now, calculating the value of 'k' (rate constant) by using any expression.

1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})

k=60.4M^{-1}s^{-1}

Now we have to calculate the rate for a reaction when concentration of OCl^-  and I^-  is 1.8\times 10^{-3}M and 6.0\times 10^{-4}M respectively.

\text{Rate}=k[OCl^-][I^-]

\text{Rate}=(60.4M^{-1}s^{-1})\times (1.8\times 10^{-3}M)(6.0\times 10^{-4}M)

\text{Rate}=6.52\times 10^{-5}Ms^{-1}

Therefore, the rate of the reaction is 6.52\times 10^{-5}Ms^{-1}

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3 years ago
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