Login/Sign Up
Home » Based on thermochemical conditions (a), (b) and (c), discover which of the logarithmic connections given in alternatives (I) to (iv) is right.    
Based on thermochemical conditions (a), (b) and (c), discover which of the logarithmic connections given in alternatives (I) to (iv) is right.

    \[\begin{array}{*{35}{l}} \left( a \right)\text{ }C\text{ }\left( graphite \right)\text{ }+\text{ }O2\text{ }\left( g \right)\text{ }\to \text{ }CO2\text{ }\left( g \right)\text{ };\text{ }rH\text{ }=\text{ }x\text{ }kJ\text{ }mol1 \\ ~ \\ \left( b \right)\text{ }C\text{ }\left( graphite \right)\text{ }+12\text{ }O2\text{ }\left( g \right)\text{ }\to \text{ }CO\text{ }\left( g \right)\text{ };\text{ }rH\text{ }=\text{ }y\text{ }kJ\text{ }mol1 \\ ~ \\ \left( c \right)\text{ }CO\text{ }\left( g \right)\text{ }+12\text{ }O2\text{ }\left( g \right)\text{ }\to \text{ }CO2\text{ }\left( g \right)\text{ };\text{ }rH\text{ }=\text{ }z\text{ }kJ\text{ }mol1 \\ ~ \\ \left( I \right)\text{ }z\text{ }=\text{ }x\text{ }+\text{ }y \\ ~ \\ \left( ii \right)\text{ }x\text{ }=\text{ }y\text{ }\text{ }z \\ ~ \\ \left( iii \right)\text{ }x\text{ }=\text{ }y\text{ }+\text{ }z \\ ~ \\ \left( iv \right)\text{ }y\text{ }=\text{ }2z\text{ }\text{ }x \\ \end{array}\]

CBSE | Chemistry | Class 11 | Exercise 1 | NCERT Exemplar | Thermodynamics
Based on thermochemical conditions (a), (b) and (c), discover which of the logarithmic connections given in alternatives (I) to (iv) is right.

    \[\begin{array}{*{35}{l}} \left( a \right)\text{ }C\text{ }\left( graphite \right)\text{ }+\text{ }O2\text{ }\left( g \right)\text{ }\to \text{ }CO2\text{ }\left( g \right)\text{ };\text{ }rH\text{ }=\text{ }x\text{ }kJ\text{ }mol1 \\ ~ \\ \left( b \right)\text{ }C\text{ }\left( graphite \right)\text{ }+12\text{ }O2\text{ }\left( g \right)\text{ }\to \text{ }CO\text{ }\left( g \right)\text{ };\text{ }rH\text{ }=\text{ }y\text{ }kJ\text{ }mol1 \\ ~ \\ \left( c \right)\text{ }CO\text{ }\left( g \right)\text{ }+12\text{ }O2\text{ }\left( g \right)\text{ }\to \text{ }CO2\text{ }\left( g \right)\text{ };\text{ }rH\text{ }=\text{ }z\text{ }kJ\text{ }mol1 \\ ~ \\ \left( I \right)\text{ }z\text{ }=\text{ }x\text{ }+\text{ }y \\ ~ \\ \left( ii \right)\text{ }x\text{ }=\text{ }y\text{ }\text{ }z \\ ~ \\ \left( iii \right)\text{ }x\text{ }=\text{ }y\text{ }+\text{ }z \\ ~ \\ \left( iv \right)\text{ }y\text{ }=\text{ }2z\text{ }\text{ }x \\ \end{array}\]

solution:

 

Choice (iii) is the appropriate response.

    \[\begin{array}{*{35}{l}} \left( a \right)\text{ }C\text{ }\left( graphite \right)\text{ }+\text{ }O2\text{ }\left( g \right)\text{ }\to \text{ }CO2\text{ }\left( g \right)\text{ };\text{ }rH\text{ }=\text{ }x\text{ }kJ\text{ }mol1\text{ }\ldots \text{ }.\left( 1 \right) \\ ~ \\ \left( b \right)\text{ }C\text{ }\left( graphite \right)\text{ }+12\text{ }O2\text{ }\left( g \right)\text{ }\to \text{ }CO\text{ }\left( g \right)\text{ };\text{ }rH\text{ }=\text{ }y\text{ }kJ\text{ }mol1\text{ }\ldots \text{ }.\left( 2 \right) \\ ~ \\ \end{array}\]

Substracting condition 1 and 2 we get

 

    \[\left( c \right)\text{ }CO\text{ }\left( g \right)\text{ }+12\text{ }O2\text{ }\left( g \right)\text{ }\to \text{ }CO2\text{ }\left( g \right)\text{ };\text{ }rH\text{ }=\text{ }z\text{ }kJ\text{ }mol1\text{ }\ldots \text{ }.\left( 3 \right)\]

Condition 3 is gotten by taking away (2) from (1)

 

In this way ,

 

    \[x\text{ }=\text{ }y\text{ }+\text{ }z\]

i

More Material