Question

predict whether the following reactions will be exothermic or endothermic.
reaction a. $n_2(g)+3h_2(g)\to2nh_3(g)$
reaction b. $s(g)+o_2(g)\to so_2(g)$
reaction c. $2h_2o(g)\to2h_2(g)+o_2(g)$
reaction d. $2f(g)\to f_2(g)$
which reaction(s) are endothermic?
b
a
d
c
which reaction(s) are exothermic?
a
c
d
b

Explanation:

Step1: Calculate $\Delta H^{\circ}$ for Reaction A

• Bond breaking (reactants):
• $1$ mole of $N\equiv N$ bonds: energy absorbed = $942$ kJ
• $3$ moles of $H - H$ bonds: energy absorbed = $3\times432=1296$ kJ
• Total energy absorbed for bond breaking: $942 + 1296=2238$ kJ
• Bond forming (products):
• $2$ moles of $NH_3$. Each $NH_3$ has $3$ $N - H$ bonds. So, $2\times3 = 6$ moles of $N - H$ bonds
• Energy released for bond forming: $6\times386 = 2316$ kJ
• $\Delta H^{\circ}=2238-2316=-78$ kJ/mol (exothermic)

Step2: Calculate $\Delta H^{\circ}$ for Reaction B

• Bond breaking (reactants):
• $1$ mole of $O = O$ bonds: energy absorbed = $494$ kJ
• Bond forming (products):
• $SO_2$ has $2$ $S = O$ bonds. Energy released for bond forming: $2\times522 = 1044$ kJ
• $\Delta H^{\circ}=494-1044=-550$ kJ/mol (exothermic)

Step3: Calculate $\Delta H^{\circ}$ for Reaction C

• Bond breaking (reactants):
• $2$ moles of $H_2O$. Each $H_2O$ has $2$ $H - O$ bonds. So, $2\times2=4$ moles of $H - O$ bonds
• Energy absorbed for bond breaking: $4\times459 = 1836$ kJ
• Bond forming (products):
• $2$ moles of $H - H$ bonds: energy released = $2\times432 = 864$ kJ
• $1$ mole of $O = O$ bonds: energy released = $494$ kJ
• Total energy released for bond forming: $864+494 = 1358$ kJ
• $\Delta H^{\circ}=1836 - 1358=478$ kJ/mol (endothermic)

Step4: Calculate $\Delta H^{\circ}$ for Reaction D

• Bond breaking (reactants):
• No bonds to break (atomic $F$)
• Bond forming (products):
• $1$ mole of $F - F$ bonds: energy released = $155$ kJ
• $\Delta H^{\circ}=0 - 155=-155$ kJ/mol (exothermic)

Answer:

• Endothermic reaction: C
• Exothermic reactions: A, B, D