CO CO2 N2O evam N2O4 ki virchan enthalpy kramsah −110.-393.81.or 9.7 kj/mol hai abhikriya N2O4+3CO2 ke liye…

Question

CO CO2 N2O evam N2O4 ki virchan enthalpy kramsah −110.-393.81.or 9.7 kj/mol hai abhikriya N2O4+3CO2 ke liye ∆rH ka maan gyat kijiye

ScanSolve AI Tutor · Accepted Answer

The problem asks to calculate the enthalpy change ( $\Delta_r H$ ) for the reaction $\ce{N2O4 + 3CO -> N2O + 3CO2}$ .
To calculate $\Delta_r H$ , we need the standard enthalpy of formation ( $\Delta_f H^\circ$ ) for each reactant and product. The formula is:

\Delta_r H^\circ = \sum n_p \Delta_f H^\circ (\text{products}) - \sum n_r \Delta_f H^\circ (\text{reactants})

The compounds involved in the reaction are $\ce{N2O4}$ , $\ce{CO}$ , $\ce{N2O}$ , and $\ce{CO2}$ .
The problem states that the formation enthalpies for $\ce{CO}$ , $\ce{CO2}$ , $\ce{N2O}$ , and $\ce{N2O4}$ are respectively $-110$ ,
$-393.81$ , and $9.7\ \text{kJ/mol}$ .

Identify given enthalpy of formation values
The given standard enthalpies of formation ( $\Delta_f H^\circ$ ) are:
$\Delta_f H^\circ (\ce{CO}) = -110\ \text{kJ/mol}$
$\Delta_f H^\circ (\ce{CO2}) = -393.81\ \text{kJ/mol}$
$\Delta_f H^\circ (\ce{N2O}) = 9.7\ \text{kJ/mol}$
Identify missing enthalpy of formation value
The problem statement lists four compounds ( $\ce{CO}$ , $\ce{CO2}$ , $\ce{N2O}$ , and $\ce{N2O4}$ ) but provides only three enthalpy of formation values. Based on the order given, the standard enthalpy of formation for $\ce{N2O4}$ is missing.
Conclusion
Since the standard enthalpy of formation for $\ce{N2O4}$ is not provided, the enthalpy change ( $\Delta_r H$ ) for the reaction $\ce{N2O4 + 3CO -> N2O + 3CO2}$ cannot be calculated.

CO CO2 N2O evam N2O4 ki virchan enthalpy kramsah -110.-393.81.or 9.7 kj/mol hai abhikriya N2O4+3CO2 ke liye ∆rH ka maan gyat kijiye

Step-by-step solution

Identify given enthalpy of formation values

Identify missing enthalpy of formation value

Conclusion

Still stuck on this one?

More Chemistry Questions