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Given N2(g) + 3H2(g) gives 2NH3(g) ; Delta r H = minus 92.4 kJ mol raised to minus 1 What is the standard enthalpy of formation of NH3 gas?

6.13  Given  N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) ;\Delta rH = -92.4 kJ mol ^{-1}.What is the standard enthalpy of formation of NH_3 gas?

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M manish

the standard enthalpy of formation of any compound is the required change in enthalpy of formation of 1 mole of a substance in its standard form from its constituent elements.

Thus we can re-write the reaction as;
\frac{1}{2}N_{2}+\frac{3}{2}H_{2}\rightarrow NH_{3}

Therefore, standard enthalpy of formation of ammonia is = \frac{1}{2}\Delta _rH^\Theta
                                                                                    = 1/2 (-92.4)
                                                                                    = -46.2 kJ/mol

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