Chemistry, 30.08.2019 22:30 sl3olsonow8388

The formation of nitryl fluoride, 2no2(g)+f2(g)⇌2no2f(g) is first order in no2 and first order in f2. if the concentration of no2 was increased by half and the concentration of f2 was quadrupled, by what factor would the reaction rate increase?

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The reaction is:

2 NO₂ (g) + F₂ (g) ⇆ 2 NO₂F (g)

The stoichiometric coefficients of the substances balance out each other to obey the Law of Definite Proportions. Now, you have to note that determining the reaction rate expression is specific to a certain type of reaction. So, this are determined empirically through doing experiments. But in chemical reaction engineering, to make things simple, you assume that the reaction is elementary. This means that the order of a reaction with respect to a certain substance follows their individual stoichiometric coefficients. What I'm saying is, the stoichiometric coefficients are the basis of our reaction rate orders. For this reaction, the rate order is 2 for NO₂, 1 for F₂ and 2 for NO₂F. When the forward and reverse reactions are in equilibrium, then it applies that:

Reaction rate of disappearance of reactants = Reaction rate of formation of products.

Therefore, we can have two reaction rate constants for this. But since the conditions manipulated are the reactant side, let's find the expression for reaction rate of disappearance of reactants.

-r = k[NO₂]²[F₂]

The negative sign before r signifies the rate of disappearance. If it were in terms of the product, that would have been positive. The term k denotes for the reaction rate constant. That is also empirical. As you can notice the stoichiometric coefficients are exponents of the concentrations of the reactants. Let's say initially, there are 1 M of NO₂ and 1 M of F₂. Then,

-r = k(1)²(1)
-r = k

Now, if we change 1 M of NO₂ by increasing it to its half, it would now be 1.5 M NO₂. Then, if we quadruple the concentration of F₂, that would be 4 M F₂. Substituting the values:

-r = k(1.5)²(4)
-r = 9k

So, as you can see the reaction rate increase by a factor of 9.

Answer from: Quest

periodic trends is an atom that gets larger as the number of electronic shells increase, the radius of atoms increases as you go down a certain group in the periodic table of elements. in general, the size of an atom will decrease as you move from left to right of a certain period.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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The formation of nitryl fluoride, 2no2(g)+f2(g)⇌2no2f(g) is first order in no2 and first order in f2...

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