Calculate the standard enthalpy change (in kJ m o l - 1 ) for the reaction, H 2 ( g ) + O 2 ( g ) → H 2 O 2 ( g ) , given, that bond enthalpies of H – H , O = O , O – H and O – O (in k J m o l - 1 ) are respectively 438, 498, 464 and 138

Question

Calculate the standard enthalpy change (in kJ m o l - 1 ) for the reaction, H 2 ( g ) + O 2 ( g ) → H 2 O 2 ( g ) , given, that bond enthalpies of H – H , O = O , O – H and O – O (in k J m o l - 1 ) are respectively 438, 498, 464 and 138, - 334, - 130, + 334, + 130

MARKS App · Accepted Answer

H 2 ( g ) + O 2 ( g ) → H 2 O 2 ( g ) Δ H r e a c t i o n = B . E . R e a c t a n t s - B . E . p r o d u c t s = [ B . E . ( H - H ) + B . E . ( O = O ) ] - [ 2 B . E . ( O - H ) + B . E . ( O - O ) ] = 438 + 498 – 2 × 464 + 138 = 936 – 1066 = - 130 k J m o l - 1

Search any question & find its solution

Looking for more such questions to practice?