Question

use the following information to answer the next question.
given the following data:
n₂h₄(l) + ch₄o(l) → ch₂o(g) + n₂(g) + 3h₂(g) δ,h = -37 kj
n₂(g) + 3h₂(g) → 2nh₃(g) δ,h = -91.8 kj
ch₄o(l) → ch₂o(g) + h₂(g) δ,h = -65 kj
using hesss law, the enthalpy of reaction for the following equation: n₂h₄(l) + h₂(g) → 2nh₃(g) is calculated to be kj.
record your 2 - digit answer. include sign in the first box only; do not include units in second box.

Explanation:

Step1: Write down the target reaction

We want to find the enthalpy for \(N_{2}H_{4}(l)+H_{2}(g)\to2NH_{3}(g)\)

Step2: Manipulate the given reactions

• Given reaction 1: \(N_{2}H_{4}(l)+CH_{4}O(l)\to CH_{2}O(g)+N_{2}(g)+3H_{2}(g)\quad\Delta_{r}H = - 37\space kJ\)
• Given reaction 2: \(N_{2}(g)+3H_{2}(g)\to2NH_{3}(g)\quad\Delta_{r}H=-91.8\space kJ\)
• Given reaction 3: \(CH_{4}O(l)\to CH_{2}O(g)+H_{2}(g)\quad\Delta_{r}H=-65\space kJ\). Reverse reaction 3: \(CH_{2}O(g)+H_{2}(g)\to CH_{4}O(l)\quad\Delta_{r}H = 65\space kJ\)

Step3: Combine the reactions

Add reaction 1, reaction 2 and the reversed reaction 3:

$$
LATEXBLOCK0
$$

Cancel out \(CH_{4}O(l)\), \(CH_{2}O(g)\), \(N_{2}(g)\) and \(3H_{2}(g)\) on both sides.

Step4: Calculate the enthalpy change

\(\Delta_{r}H=(-37)+(-91.8)+65\)

$$
LATEXBLOCK1
$$

Answer:

-64