Question

Calculate the standard enthalpy of the reaction.
∆ CO(g) + 2NO2(g) —> 4CO2(g) + N2(g)
If, ∆fH° (CO) = -110.5 KJ/Mol
∆fH° (CO2) = -393.5 KJ/ Mol
∆fH° (NO2) = 33.2 KJ/Mol

Only for class 11th..!!
pls calculation ke saath batao!!​

Answer 1

Answer:

–1529.9 KJ/mol

Explanation:

dfH°= 4×dfH°(CO2)+dfH°(N2)–dfH°(CO)–2×dfH°(NO2)

= 4×(–393.5)+0–(–110.5)–2×(33.2)

= –1529.9 KJ/ mol