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The IP(1) of Li is 5.41eV and E.A of Cl ...

The `IP_(1)` of Li is 5.41eV and E.A of Cl is -3.61eV. The `DeltaH` in KJ/mole for the reaction `Li(g) + Cl(g) rarr Li^(+)(g) +Cl^(-)(g)` is `1.736 xx 10^(x)` KJ/mole. The value of 'x' is

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The first IP lithium is 5.41eV and electron gain enthalpy of Cl is -3.61eV . Calculate Delta H in KJ mol^(-1) for the reaction: Li_(g) +Cl_(g) rarr Li_(g)^= +Cl_(g)^(-) .

If the first ionization enthalpy of Li is 5.4 eV and the elec- tron gain enthalpy of chlorine is 3.6 eV then the Delta H in kcal/mole for the reaction will be Li(g)+Cl(g) to Li^(+) (g) Cl^(-) (g) (Presume that the pressure is so low that the ions do not combine with each other)

The first ionisation potential of Li is 5.4 e V and the electron affinity of Cl is 3.6 eV . Calculate Delta H in kcal mol^(-1) for the reaction Li (g) + Cl (g) to Li^(+) + Cl^(-) Carried out at such low pressures that resulting ions do not combine with each other.

For the reaction, 2Cl(g) rarr Cl_2(g) , the correct option is:

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