QuestionJuly 30, 2025

Calculate the Delta H for the reaction NO_((g))+O_((g))arrow NO_(2(g)) Given the following information:Eq 1:NO_((g))+O_(3(g))arrow NO_(2(g))+O_(2(g)) Delta H=-198.9kJ/mol Eq 2: O_(3(g))arrow 3/2O_(2(g)) Delta H=-142.3kJ/mol Eq 3: O_(2(g))arrow 2O_((g)) Delta H=495.0kJ/mol

Calculate the Delta H for the reaction NO_((g))+O_((g))arrow NO_(2(g)) Given the following information:Eq 1:NO_((g))+O_(3(g))arrow NO_(2(g))+O_(2(g)) Delta H=-198.9kJ/mol Eq 2: O_(3(g))arrow 3/2O_(2(g)) Delta H=-142.3kJ/mol Eq 3: O_(2(g))arrow 2O_((g)) Delta H=495.0kJ/mol
Calculate the Delta H for the reaction	NO_((g))+O_((g))arrow NO_(2(g))
Given the following information:Eq 1:NO_((g))+O_(3(g))arrow NO_(2(g))+O_(2(g)) Delta H=-198.9kJ/mol
Eq 2: O_(3(g))arrow 3/2O_(2(g)) Delta H=-142.3kJ/mol
Eq 3: O_(2(g))arrow 2O_((g)) Delta H=495.0kJ/mol

Solution
4.1(251 votes)

Answer

\( \Delta \mathrm{H} = -836.2 \, \mathrm{kJ/mol} \) Explanation 1. Reverse Eq 3 Reverse Eq 3 to match the formation of \( \mathrm{O}_{2(\mathrm{g})} \) from \( \mathrm{O}_{(\mathrm{g})} \). New \( \Delta \mathrm{H} = -495.0 \, \mathrm{kJ/mol} \). 2. Combine Eqs 1 and 2 Add Eq 1 and Eq 2 to eliminate \( \mathrm{O}_{3(\mathrm{g})} \), resulting in \( \mathrm{NO}_{(\mathrm{g})} + \frac{3}{2} \mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{NO}_{2(\mathrm{g})} + \mathrm{O}_{2(\mathrm{g})} \). Total \( \Delta \mathrm{H} = -198.9 \, \mathrm{kJ/mol} + (-142.3 \, \mathrm{kJ/mol}) = -341.2 \, \mathrm{kJ/mol} \). 3. Add reversed Eq 3 Add the reversed Eq 3 to the result of Step 2 to form \( \mathrm{NO}_{(\mathrm{g})} + \mathrm{O}_{(\mathrm{g})} \rightarrow \mathrm{NO}_{2(\mathrm{g})} \). Total \( \Delta \mathrm{H} = -341.2 \, \mathrm{kJ/mol} + (-495.0 \, \mathrm{kJ/mol}) = -836.2 \, \mathrm{kJ/mol} \).

Explanation

1. Reverse Eq 3<br /> Reverse Eq 3 to match the formation of \( \mathrm{O}_{2(\mathrm{g})} \) from \( \mathrm{O}_{(\mathrm{g})} \). New \( \Delta \mathrm{H} = -495.0 \, \mathrm{kJ/mol} \).<br />2. Combine Eqs 1 and 2<br /> Add Eq 1 and Eq 2 to eliminate \( \mathrm{O}_{3(\mathrm{g})} \), resulting in \( \mathrm{NO}_{(\mathrm{g})} + \frac{3}{2} \mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{NO}_{2(\mathrm{g})} + \mathrm{O}_{2(\mathrm{g})} \). Total \( \Delta \mathrm{H} = -198.9 \, \mathrm{kJ/mol} + (-142.3 \, \mathrm{kJ/mol}) = -341.2 \, \mathrm{kJ/mol} \).<br />3. Add reversed Eq 3<br /> Add the reversed Eq 3 to the result of Step 2 to form \( \mathrm{NO}_{(\mathrm{g})} + \mathrm{O}_{(\mathrm{g})} \rightarrow \mathrm{NO}_{2(\mathrm{g})} \). Total \( \Delta \mathrm{H} = -341.2 \, \mathrm{kJ/mol} + (-495.0 \, \mathrm{kJ/mol}) = -836.2 \, \mathrm{kJ/mol} \).
Click to rate:

Similar Questions