Question

Use the bond enthalpy data in Table 7.67.6 to calculate the enthalpies of formation of the following gases:
H2NCl\mathrm{H}_2 \mathrm{NCl}.

Solution

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From the reaction of chloramine formation we get:

H2(g)+12N2(g)+12Cl2(g)NH2Cl(g)\mathrm{H_2(g)+\dfrac{1}{2}N_2(g)+\dfrac{1}{2}Cl_2(g) \rightarrow NH_2Cl(g)}

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