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Leya [2.2K]
3 years ago
10

For the following aqueous reaction, complete and balance the molecular equation and write a net iconic equatio, making sure to i

nclude the physical states of the compounds:
Potassium carbonate + Strontium nitrate
Chemistry
1 answer:
pav-90 [236]3 years ago
6 0

Answer:

Balance molecular equation:

K2CO3(aq) + Sr(NO3)2(aq) → SrCO3(s) + 2KNO3(aq)

Net ionic equation:

CO3∧-2(aq) + Sr∧+2(aq) → SrCO3(s)

Explanation:

Potassium carbonate = K2CO3

Strontium nitrate = Sr(NO3)2

Chemical equation:

K2CO3 + Sr(NO3)2 → SrCO3 + KNO3

Balance chemical equation with physical states:

K2CO3(aq) + Sr(NO3)2(aq) → SrCO3(s) + 2KNO3(aq)

Ionic equation:

2K+(aq) + CO3∧-2(aq) + Sr∧+2(aq) + 2NO∧-3(aq) → SrCO3(s) + 2K+(aq) + 2NO∧-3(aq)

Net ionic equation:

CO3∧-2(aq) + Sr∧+2(aq) → SrCO3(s)

2K+ and 2NO∧-3 ions are spectator ions that's way these are not written in net ionic equation.

Spectator ions:

These are the ions that are present same on both side of chemical reaction and does not effect the equilibrium.

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Answer:

<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

A+B\rightarrow C+D

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

Rate\ \alpha [A]^{a}[B]^{b}

r=[A]^{a}[B]^{b}.................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

r'=[A']^{a}[B']^{b}

Put the value of [A'] , [B'] and r' in the above equation:

2r=[2A]^{a}[B]^{b}...........(2)

Divide equation (1) by (2) we , get

\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}

2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}

Here A and A cancel each other

B and B cancel each other

We get,

2= 2^{a}\times 1^{b}

1^b = 1 ( power of 1 = 1)

2= 2^{a}

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1  = 2

b = 2

Hence the rate law becomes :

r=[A]^{a}[B]^{b}

<u>r=[A]^{1}[B]^{2}.............(3)</u>

Look in the question now, it is asked to calculate the concentration of [B],if  cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

r'=[A]^{1}[B']^{2}

r'=[A]^{1}(\frac{1}{2}[B])^{2}

r'=\frac{1}{4}[A]^{1}[B]^{2}............(a)

But

r=[A]^{a}[B]^{b}..............(b)

Compare equation (a) and (b) , we get

new rate r' =

<u>r' = 1/4 r</u>

7 0
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From the equation provided, hydrogen gas is given off as a by-product

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