For the reaction 2 NO(g) + 2 H2(g) => N2(g) + 2 H2O(g) the following data were collected. What is the rate law for this react
1 answer:
<u>Answer:</u> The rate law expression for the given reaction is written below.
<u>Explanation:</u>
Rate law 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:
Rate law expression for the reaction:
Hence, the rate law expression for the given reaction is written above.
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<u>Answer:</u> The mass of produced is 2.52 g
<u>Explanation:</u>
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
......(1)
- <u>For hexane:</u>
Given mass of hexane = 1.72 g
Molar mass of hexane = 86.18 g/mol
Putting values in equation 1, we get:
- <u>For oxygen gas:</u>
Given mass of oxygen gas = 8.0 g
Molar mass of oxygen gas= 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the combustion of hexane follows:
By stoichiometry of the reaction:
If 2 moles of hexane reacts with 19 moles of oxygen gas
So, 0.020 moles of hexane will react with = of oxygen gas
As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, hexane is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 2 moles of hexane produces 14 moles of
So, 0.020 moles of hexane will produce = of
We know, molar mass of = 18 g/mol
Putting values in above equation, we get:
Hence, the mass of produced is 2.52 g
Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.
m = molality of solute (m or )
Given : kf = 1.86
m = 0.907 )
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86 * 0.907
)
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C