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3 years ago

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Consider the formation of nitryl fluoride: 2NO2(g)+F2(g)⇌2NO2F(g) The reaction is first order in F2 and second order overall. Wh

at is the rate law? View Available Hint(s) Consider the formation of nitryl fluoride: The reaction is first order in and second order overall. What is the rate law? rate=k[NO2]2[F2]2 rate=k[NO2][F2]2 rate=k[F2] rate=k[NO2] rate=k[NO2][F2] rate=k[NO2]2[F2]

1 answer:

Dahasolnce [82]3 years ago

7 0

Answer:

Rate = $k[NO_{2}][F_{2}]$

Explanation:

Two reactants are present in this reaction which are $NO_{2}and F_{2}$
We know overall order of a reaction is summation of individual order with respect to reactants present in rate law equation.
Here, overall order of reaction is 2 including first order with respect to $F_{2}$
So, rate of reaction should also be first order with respect to another reactant i.e. first order with respect to $NO_{2}$
So, rate law: rate = $k[NO_{2}][F_{2}]$

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Given the balanced equation:

Aliun [14]

Answer:

0.4

Explanation:

Given data:

Number of moles of SrCl₂ consumed = ?

Mass of ZnCl₂ produced = 54 g

Solution:

Chemical equation:

ZnSO₄ + SrCl₂ → SrSO₄ + ZnCl₂

Number of moles of ZnCl₂:

Number of moles = mass/ molar mass

Number of moles = 54 g/136.3 g/mol

Number of moles = 0.4 mol

Now we will compare the moles of ZnCl₂ with SrCl₂ from balance chemical equation.

ZnCl₂ : SrCl₂

1 : 1

0.4 : 0.4

Thus when 54 g of ZnCl₂ produced 0.4 moles of SrCl₂ react.

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3 years ago

II. Ionic Equations

mario62 [17]

Answer:

Complete ionic: $\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Net ionic: $\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

Explanation:

Start by identifying species that exist as ions. In general, such species include:

Soluble salts.
Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ . These three salts will exist as ions:

Each $\rm AgNO_3\, (aq)$ formula unit will exist as one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion.
Each $\rm CaCl_2$ formula unit will exist as one $\rm Ca^{2+}$ ion and two $\rm Cl^{-}$ ions (note the subscript in the formula $\rm CaCl_2\!$ .)
Each $\rm Ca(NO_3)_2$ formula unit will exist as one $\rm Ca^{2+}$ and two $\rm {NO_3}^{-}$ ions.

On the other hand, $\rm AgCl$ is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each $\rm AgNO_3\, (aq)$ formula unit will exist as only one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion. However, because the coefficient of $\rm AgNO_3\, (aq)\!$ in the original equation is two, $\!\rm AgNO_3\, (aq)$ alone should correspond to two $\rm Ag^{+}\!$ ions and two $\rm {NO_3}^{-}\!$ ions.

Do not rewrite the salt $\rm AgCl$ because it is insoluble.

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of $\rm Ca^{2+}$ and two units of $\rm {NO_3}^{-}$ . Doing so will give:

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}$ .

Simplify the coefficients:

$\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

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