Chemistry: Gthe following reaction is the first step in the production of nitric acid from ammonia. 4nh3(g) 5o2(g) → 4no(g) 6h2o(g) calculate the standard enthalpy change, in kj, for this reaction

Gthe following reaction is the first step in

Ответ:

Cjohnston742

10.01.2022

The enthalpy of the reaction is coming out to be -902 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(4\times \Delta H_f_{(NO(g))})+(6\times \Delta H_f_{(H_2O(g))})]-[(4\times \Delta H_f_{(NH_3(g))})+(5\times \Delta H_f_{(O_2)})]$

We are given:

$\Delta H_f_{(NO(g))}=91.3kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-45.9kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(4\times (91.3))+(6\times (-241.8))]-[(4\times (-45.9))+(5\times (0))]\\\\\Delta H_{rxn}=-902kJ$

Hence, the enthalpy of the reaction is coming out to be -902 kJ.

0,0(0 оценок)

Ответ:

neilaamusic6966

05.06.2022

Chemistry

Zero

Explanation:

The addition of all forces acting on a body which is in equilibrium must be zero, since there must'n be any acceleration (net acting force must be zero)

0,0(0 оценок)

Gthe following reaction is the first step in the production of nitric acid from ammonia. 4nh3(g) 5o2(g) → 4no(g) 6h2o(g) calculate the standard enthalpy change, in kj, for this reaction

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Chemistry

Ask an AI advisor a question