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ISC Class 12 Chemistry Solved Question Paper 2007

Short Answer Type

21. Define molecularity of a reaction. Give one difference between the order of a reaction and its molecularity.

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22. Which one of the following O₂⁺ or O ^-₂ , is considered to be more stable ? Justify your answer on the basis of the Molecular Orbital Theory.

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23. Draw the electron dot diagrams or any other suitable representation of chloric acid and hypochlorous acid.

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24.

(i) How many sodium ions and chloride ions are present in a unit cell of sodium chloride ?

(ii) What is the co-ordination number of sodium and chloride ions in sodium chloride crystals ?

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25.

Give a reason for each of the following :

(i) Specific conductance decreases with dilution whereas equivalent conductance increases with dilution.

(ii) Lead is precipitated as PbS while zinc is not precipitated when H₂S gas is passed through an acidic solution of lead nitrate and zinc nitrate.

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26. Mention any two factors that influence the rate of a chemical reaction

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27.

(i) Enthalpy change and entropy change of a chemical reaction at 25°C are 177.0 KJ and 160.4 JK^-1 respectively. Calculate free energy change of the reaction and predict whether the reaction is spontaneous or nonspontaneous.

(ii) 2 moles of an ideal gas at 27°C expand reversibly from 2 litres to 20 litres. Find the entropy change for the reaction.

(iii) State the second law of thermodynamics in terms of entropy of the universe.

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28. If enthalpies of formation for C₂H₄, C0₂ and H₂0 at 25°C and 1 atm. pressure are 52 KJ/mole, -394 KJ/mole and -286 KJ/mole respectively, calculate the enthalpy of combustion of C₂H₄.

C₂H₄ + 20₂→2 C0₂ + 2 H₂0

Given ΔH _f⁰ C₂H₄ = 52 KJ/mole

ΔH _f⁰ C0₂ = - 394 KJ/mole

ΔH_f⁰ H₂0 = -286 KJ/mole

$Enthalpy space of space combustion space space equals space increment straight H subscript straight f superscript 0 space equals space straight capital sigma increment straight H subscript straight f left parenthesis product right parenthesis end subscript superscript 0 minus space straight capital sigma increment straight H subscript straight f left parenthesis reactant right parenthesis end subscript superscript 0 therefore increment straight H subscript straight f superscript 0 space equals left square bracket 2 increment straight H subscript straight f superscript 0 CO subscript 2 space plus space 2 increment straight H subscript straight f superscript 0 space straight H subscript 2 straight O right square bracket minus left square bracket increment straight H subscript straight f superscript 0 space straight C subscript 2 straight H subscript 4 space plus 2 increment straight H subscript straight f superscript 0 space straight O subscript 2 right square bracket equals space left square bracket left parenthesis 2 space straight x space minus 394 right parenthesis space plus left parenthesis 2 space straight x space minus 286 right parenthesis right square bracket minus left square bracket 52 plus straight Z left parenthesis 2 straight x 0 right parenthesis right square bracket equals space left parenthesis negative 788 minus 572 right parenthesis minus 52 equals negative 788 minus 572 minus 52 equals negative 1412 KJ divided by mol$