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JEE Main 20231 FebruaryMorning Shift
Exam
JEE Main
Year
2023
Date
1 February
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2023 · 1 February · Morning Shift · Chemistry · Thermodynamics (C)

At 25 ° C , the enthalpy of the following processes are given: H _...

At 25°C25 ° C, the enthalpy of the following processes are given:

H2(g)+O2(g)→2OH(g)ΔHo=78kJmol−1{H}_{2} ( g ) + {O}_{2} ( g ) \to 2OH ( g ) \Delta{H}^{o} = 78kJ{mol}^{- 1}

H2(g)+1/2O2(g)→H2O(g)ΔH0=−242kJmol−1{H}_{2} ( g ) + 1 / 2{O}_{2} ( g ) \to {H}_{2}O ( g ) \Delta{H}^{0} = - 242kJ{mol}^{- 1}

H2(g)→2H(g)ΔHo=436kJmol−1{H}_{2} ( g ) \to 2H ( g ) \Delta{H}^{o} = 436kJ{mol}^{- 1}

1/2O2(g)→O(g)ΔH0=249kJmol−11 / 2{O}_{2} ( g ) \to O ( g ) \Delta{H}^{0} = 249kJ{mol}^{- 1}

What would be the value of X for the following reaction? (Nearest integer)

H2O(g)→H(g)+OH(g)∆Ho=XkJmol−1{H}_{2}O\left(g\right) \to H\left(g\right) + OH\left(g\right) ∆ {H}^{o} = X{kJmol}^{- 1}

Correct answer

0

Step-by-step solution

2H2O(g)→2H2(g)+O2(g)+(242×2)kJmol−12{H}_{2}O ( g ) \to 2{H}_{2} ( g ) + {O}_{2} ( g ) + ( 242 \times 2 ) {kJmol}^{- 1} ....(1)

H2(g)+O2(g)→2OH+78kJmol−1{H}_{2} ( g ) + {O}_{2} ( g ) \to 2OH + 78kJ{mol}^{- 1} ....(2)

H2(g)→2H(g)+436kJmol−1{H}_{2} ( g ) \to 2H ( g ) + 436kJ{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:

⇒H2O(g)→H(g)+OH(g)998×12=+499kJmol−1⇒ {H}_{2}O ( g ) \to H ( g ) + OH ( g ) 998 \times \dfrac{1}{2} = + 499kJ{mol}^{- 1}
Thus X=499X = 499

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