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3 years ago

What volume of 0.160 m li2s solution is required to completely react with 130 ml of 0.160 m co no3 2?

1 answer:

3 0

The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below

write the reacting equation

Co(NO3)2 + Li2S = 2LiNO3 + COS

find the moles of CO(NO3)2 = molarity x volume

= 130 ml x 0.160=20.8 moles

since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles

volume of Lis is therefore = moles of Lis/ molarity of LiS

= 20.8/0.160 = 130 Ml

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Diamond is forever" is one of the most successful advertising slogans of all time. but is it true? for the reaction shown below,

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The given reaction is

$C _{(Diamond)}\rightarrow C_{(Graphite)}$

An element can exist in 2 or more different forms which have totally different chemical and physical properties. They are known an allotropes.

Diamond and Graphite are allotropic forms of carbon. In the given reaction, diamond is changing to graphite and we have to find out the standard free energy of this reaction. This will help us to find out whether this reaction is spontaneous at 298 K or not.

The following data is needed for calculations which is taken from standard reference table.

$H_{f}^{0}(diamond)= 1.895 kJ/mol$

$H_{f}^{0}(graphite)= 0$

$S^{0}(diamond)=2.337 J/mol-K$

$S^{0}(graphite)=5.740 J/mol-K$

Step 1: Find ΔH⁰ rxn for the given reaction.

The formula to calculate ΔH⁰ rxn is given below.

$\bigtriangleup H^{0}_{rxn}= H_{f}(product)- H_{f}( reactant)$

We have graphite on product side and diamond on reactant side.

Therefore, $\bigtriangleup H^{0}_{rxn}= H_{f}(graphite)- H_{f}(diamond)$

Let us plug in the values given above.

$\bigtriangleup H^{0}_{rxn}= 0 - 1.895 kJ/mol$

$\bigtriangleup H^{0}_{rxn}= - 1.895 kJ/mol$

ΔH⁰ rxn for the given reaction is -1.895 kJ/mol

Step 2 : Find ΔS⁰ rxn for the given reaction.

The formula to calculate ΔS⁰ rxn is

$\bigtriangleup S^{0}_{rxn}= S^{0}(product)- S^{0}( reactant)$

$\bigtriangleup S^{0}_{rxn}= S^{0}(graphite)- S^{0}(diamond)$

$\bigtriangleup S^{0}_{rxn}= (5.740J/mol.K) - (2.337 J/mol.K)$

$\bigtriangleup S^{0}_{rxn}= 3.403 J/mol.K$

Let us convert this to kJ.

$\frac{3.403J}{mol.K}\times \frac{1 kJ}{1000J}= 3.403\times 10^{-3}kJ/mol.K$

ΔS⁰ rxn for the given reaction is 3.403 x 10⁻³ kJ/mol.K

Step 3: Find standard free energy ΔG⁰ rxn.

ΔG⁰ rxn for the given reaction is calculated as

$\bigtriangleup G^{0}_{rxn}= \bigtriangleup H^{0}_{rxn}- T\times \bigtriangleup S^{0}_{rxn}$

We have T = 298 K. Let us plug in the calculated values of ΔH⁰ rxn and ΔS⁰ rxn.

$\bigtriangleup G^{0}_{rxn}= - 1.895 kJ/mol - [(298K)\times 3.403\times 10^{-3}kJ/mol.K]$

$\bigtriangleup G^{0}_{rxn}= - 1.895 kJ/mol - [1.014 kJ/mol]$

$\bigtriangleup G^{0}_{rxn}= - 2.909 kJ/mol$

The standard free energy change for the given reaction is -2.909 kJ/mol

The negative value of delta G⁰ suggests that the given reaction is spontaneous at room temperature. That means diamond will slowly convert to graphite. The speed of this reaction is extremely slow, but yet the reaction is taking place. So over a period of time diamond will become graphite.

Therefore "Diamond is forever" is not true as it is going to get converted to graphite.

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3 years ago