Question

For the reaction A → B; ∆H = + 24 kJ/mol and B → C;
∆H = – 18 kJ/mol, the decreasing order of enthalpy of A, B
and C follows the order
(a) A, B, C (b) B, C, A
(c) C, B, A (d) C, A, B

Answer 1

answer : option (b) B, C , A

given, A ⇒B ; ∆H = +24 kJ/mol

means, heat of products - heat of reactants = +24 kJ/mol

⇒H_B - H_A = +24 kJ/mol.......(1)

it is clear that, H_B > H_A

again, B ⇒C ; ∆H = -18 kJ/mol

heat of products - heat of reactants = -18

⇒H_C - H_B = -18 kJ/mol........(2)

it is clear that, H_B > H_C

from equations (1) and (2) we get,

H_C - H_A = +6 kJ/mol and hence, H_C > H_A

so, it is clear that , H_B > H_C > H_A

hence option (b) is correct choice.

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