At 25 ° C , the enthalpy of the following processes are given: H 2 ( g ) + O 2 ( g ) → 2 OH ( g ) Δ H o = 78 kJ mol - 1 H 2 ( g ) + 1 / 2 O 2 ( g ) → H 2 O ( g ) Δ H 0 = - 242 kJ mol - 1 H 2 ( g ) → 2

Question

At 25 ° C , the enthalpy of the following processes are given: H 2 ( g ) + O 2 ( g ) → 2 OH ( g ) Δ H o = 78 kJ mol - 1 H 2 ( g ) + 1 / 2 O 2 ( g ) → H 2 O ( g ) Δ H 0 = - 242 kJ mol - 1 H 2 ( g ) → 2 H ( g ) Δ H o = 436 kJ mol - 1 1 / 2 O 2 ( g ) → O ( g ) Δ H 0 = 249 kJ mol - 1 What would be the value of X for the following reaction? (Nearest integer) H 2 O g → H g + OH g ∆ H o = X kJmol - 1

TestHub · Accepted Answer

2 H 2 O ( g ) → 2 H 2 ( g ) + O 2 ( g ) + ( 242 × 2 ) kJmol - 1 ....(1) H 2 ( g ) + O 2 ( g ) → 2 OH + 78 kJ mol - 1 ....(2) H 2 ( g ) → 2 H ( g ) + 436 kJ mol - 1 ....(3) Using Hess's law of constant heat summation, adding equations(1), (2) and (3) and dividing the net result by 2 we get: ⇒ H 2 O ( g ) → H ( g ) + OH ( g ) 998 × 1 2 = + 499 kJ mol - 1 Thus X = 499

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