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Q.

For the reaction A→B; $∆$ H = +24 kJ / mol and B→C ; $∆$ H = −18 kJ/mol, the decreasing order of enthalpy of A, B, C follows the order

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a

A, B, C

b

B, C, A

c

C, A, B

d

C, B, A

answer is B.

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Detailed Solution

H_B -H_A = 24 kJ

H_B = 24 + H_A

H_C - H_B = -18 kJ

H_B = H_C + 18

H_C = 6 + H_A

H_B > H_C > H_A

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