For the reaction- A- B- H-24kJ-mole- For the reaction- B- C- H-18kJ-mole- The decreasing order of enthalpy of A- B- C follows the order-

The correct option is B B,C,A A→ B,Δ H=24kL/mole i.e., H _ B H _ A =24kJ/mole...(i) B→ C,Δ H= 18kJ/mole i.e., H _ C H _ B = 18kJ/m ...

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Byju's Answer

Standard XII

Chemistry

Enthalpy

For the react...

Question

For the reaction, $A \to B; Δ H = + 24 k J / m o l e$ . For the reaction, $B \to C, Δ H = - 18 k J / m o l e$ . The decreasing order of enthalpy of $A, B, C$ follows the order:

A, B, C

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B, C, A

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C, N, A

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C, A, B

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Solution

The correct option is B $B, C, A$
$A \to B, Δ H = 24 k L / m o l e$
i.e., $H_{B} - H_{A} = 24 k J / m o l e . . . (i)$

$B \to C, Δ H = - 18 k J / m o l e$
i.e., $H_{C} - H_{B} = - 18 k J / m o l e . . . (i i)$

Adding eqn (i) and (ii), we get-

$H_{C} - H_{A} = 6 k J / m o l e . . . (i i i)$

From eqs. (i), (ii) and (iii)-

$H_{B} > H_{C} > H_{A}$

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For the reaction, A→B;ΔH=+24kJ/mole. For the reaction, B→C,ΔH=−18kJ/mole. The decreasing order of enthalpy of A,B,C follows the order:

The correct option is B B,C,AA→B,ΔH=24kL/molei.e., HB−HA=24kJ/mole...(i)B→C,ΔH=−18kJ/molei.e., HC−HB=−18kJ/mole...(ii)Adding eqn (i) and (ii), we get-HC−HA=6kJ/mole...(iii)From eqs. (i), (ii) and (iii)-HB>HC>HA

For the reaction, $A \to B; Δ H = + 24 k J / m o l e$ . For the reaction, $B \to C, Δ H = - 18 k J / m o l e$ . The decreasing order of enthalpy of $A, B, C$ follows the order: