giúp giải bài này với mọi người :<<

$43,$

$a,$

$CH_4(g)+2O_2(g) \to CO_2(g) +2H_2O(l)$

$ΔH^o_r=ΔD_{Đứt}-ΔD_{Lắp}$

$=4D_{C-H}+2 \cdot ΔD_{O_2}-2 \cdot ΔD_{C=O/CO_2}-2 \cdot 2ΔD_{O-H}$

$=4 \cdot 413+2 \cdot 498- 2 \cdot 799-2 \cdot  2 \cdot 467$

$=-818(kJ)$

$b,$

$CH_4(g)+3Cl_2(g) \to 3HCl(g)+CHCl_3(g)$

$ΔH^o_r=ΔD_{Đứt}-ΔD_{Lắp}$

$=4D_{C-H}+3 \cdot ΔD_{Cl-Cl}-3 \cdot ΔD_{H-Cl}- ΔD_{C-H}-3ΔD_{C-Cl}$

$=4 \cdot 413+3 \cdot 243-3 \cdot 427-413-3 \cdot 339$

$=-330(kJ)$

$44,$

$a,$

$N_2(g)+3H_2(g) \to 2NH_3(g)$

$ΔH^o_r=ΔD_{Đứt}-ΔD_{Lắp}$

$=D_{N \equiv N}+3 \cdot ΔD_{H-H}-2 \cdot 3 \cdot ΔD_{N-H}$

$=945+3 \cdot 432-2 \cdot 3 \cdot 391$

$=-105(kJ)$

$b,$

$C_2H_4(g)+HBr(g) \to C_2H_5Br(g)$

$ΔH^o_r=ΔD_{Đứt}-ΔD_{Lắp}$

$=D_{C-C}+4D_{C-H}+ΔD_{H-Br}-D_{C-C}- ΔD_{C-Br}-5ΔD_{C-H}$

$=ΔD_{H-Br}- ΔD_{C-Br}-ΔD_{C-H}$

$=363-276-413$

$=-326(kJ)$

`-`$\texttt{According }$ to the table of average bond energies values ​​in the image below.