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Thermodynamics

Question
CBSEENCH11008339
Wired Faculty App

Consider the following processes Δ H (kJ/mol)

1/2 A →                               +150
3B   → 2 C + D                     -125
E + A  → 2D                        +350

For  B + D   → E + 2C, ΔH will be 

  • 525 kJ/mol

  • -175 kJ/mol

  • -325 kJ /mol

  • 325 kJ/mol

Solution

B.

-175 kJ/mol

1 half space straight A space space rightwards arrow space straight B semicolon space space space space space space space space space space space space space space space space space space space space space space increment space equals space 150 space space kJ divided by space mol space space.. space left parenthesis straight i right parenthesis 3 straight B space rightwards arrow space 2 space straight C space plus space straight D space semicolon space space space space space space space space space space space increment space straight H space equals space minus space 125 space kJ divided by mol space... space left parenthesis ii right parenthesis straight E space plus straight A space rightwards arrow space 2 straight D semicolon space space space space space space space space space space space space space space space space increment space straight H space equals space plus 350 space kJ divided by mol space.... space left parenthesis iii right parenthesis ________________________________________________ By space left square bracket space 2 space straight x space left parenthesis straight i right parenthesis space plus space left parenthesis ii right parenthesis space minus space left parenthesis iii right parenthesis comma space we space have space straight B space plus space straight D space space rightwards arrow space straight E space plus space thin space 2 straight C therefore space increment straight H space equals space 150 space straight x space 2 space space plus space left parenthesis negative 125 right parenthesis space minus space 350 equals space minus 175 space kJ divided by mol

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