Question

question 29
1 pts
calculate the standard enthalpy change for the formation of 2 moles of no(g) from 1 mole of n₂(g) and 1 mole of o₂(g)
using the following thermochemical equations:
n₂(g) + 2 o₂(g) → 2 no₂(g) δh° = 66.4 kj
2 no(g) + o₂(g) → 2 no₂(g) δh° = -114.1 kj
report your answer with three significant figures. do not include units in your answer. be sure to include a negative sign if
the answer is negative.

Explanation:

Step1: Reverse the second equation

Reverse \(2NO(g)+O_2(g)\to2NO_2(g)\), \(\Delta H^0 = 114.1\ kJ\)

Step2: Subtract the reversed second equation from the first equation

\((N_2(g)+2O_2(g)\to2NO_2(g))-(2NO(g)+O_2(g)\to2NO_2(g))\) gives \(N_2(g)+O_2(g)\to2NO(g)\)
\(\Delta H^0=66.4 - (- 114.1)\)
\(\Delta H^0=66.4 + 114.1\)
\(\Delta H^0 = 180.5\approx181\)

Answer:

181