Question

Given that bond energies of h–h and cl–cl are 430 kj mol–1 and 240 kj mol–1 respectively and ∆fh for hcl is –90 kj mol–1. bond enthalpy of hcl is : (1) 245 kj mol–1 (2) 290 kj mol–1 (3) 380 kj mol–1 (4) 425 kj mol

Answer 1

Answer:

The correct option is option (1).

Explanation:

$\frac{1}{2}H_2+\frac{1}{2}Cl_2\rightarrow HCl$

$\Delta H_{H-H}=430 kJ/mol$

$\Delta H_{Cl-Cl}=240 kJ/mol$

$\Delta H_{H-Cl}=?$

$Delta H_f=1 mol\times -90 kJ/mol=1 mol\times \Delta H_{H-Cl}-\frac{1}{2} mole\times \Delta H_{H-H}-\frac{1}{2} moles\times \Delta H_{Cl-Cl}$

$\Delta H_{H-Cl}=-90 kJ+\frac{1}{2}\times 430 kJ+\frac{1}{2}240 kJ$

$\Delta H_{H-Cl}=245 kJ$

For 1 mol HCL the $\Delta H_{H-Cl}=245 kJ/mol$

The correct option is option (1).

Answer 2

Answer:425 kJ/mol

hope it helps

Explanation:

Solution :

Given bond energies of H-H and Cl-Cl bonds which can be represented as follows ,

H2(g)→2HgΔH(H)=+430KJ/mol---------------(1)

Cl2(g)→2Cl(g)ΔHCl=+240KJ/mol------------(2)

HCl(g)→H(g)+Cl(g)ΔH(HCl)=? -------------(3)

Bond enthalpy of HCl[ΔHHCl] can be found out as follows .

ΔH=∑ΔH0f(product)−∑ΔH0f(reactantt)

=[ΔH0f(H)+ΔH0f(Cl)]−[ΔH0f(HCl)]

=1/2×430+1/2×240−(−90)

=425KJ/mol