Given N2 (g) + 3H2 (g) → 2NH3 (g) ; ∆rH0= –92.4 kJ mol–1 .What is the standard enthalpy of formation of NH3 gas?

Asked by Abhisek | 1 year ago |  66

##### Solution :-

“Standard enthalpy of formation of a compound is the enthalpy that takes place during the formation of 1 mole of a substance in its standard form, from its constituent elements in their standard state.”

Dividing the chemical equation given in the question by 2, we get

(0.5)N2(g) + (1.5)H2(g) → 2NH3(g)

Therefore, Standard Enthalpy for formation of ammonia gas

= (0.5)$$\Delta _{r}H^{\Theta }$$

= (0.5)$$(-92.4kJ mol^{-1})$$

$$-46.2kJ mol^{-1}$$

Answered by Pragya Singh | 1 year ago

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