Question

which of the following has a dh°f = 0 kj/mol?
a n₂(g)
b no(g)
c fe²⁺(aq)
d h₂o(l)
question 6
4 points
determine the amount of heat (in kj) given off when 1.26×10⁴ g of nh₃ are produced according to the equation:
n₂(g) + 3h₂(g) → 2nh₃(g) δh° = -92.6 kj/mol
a 3.4×10¹¹ kj
b 7.3×10⁻⁴ kj
c 2.1×10⁶ kj
d 3.4×10⁴ kj

Explanation:

Question 5

The standard enthalpy of formation ($\Delta H_f^{\circ}$) is defined as the change in enthalpy when one mole of a substance is formed from its elements in their standard states. Elements in their standard states have $\Delta H_f^{\circ}=0\ \text{kJ/mol}$. $N_2(g)$ is the standard state of nitrogen. $NO(g)$ is a compound (formed from $N_2$ and $O_2$), $Fe^{2 +}(aq)$ is an ion (not the standard state of iron, which is $Fe(s)$), and $H_2O(l)$ is a compound (formed from $H_2(g)$ and $O_2(g)$).

Step1: Calculate the molar mass of $NH_3$

The molar mass of $NH_3$ ($M$) is $M=(14 + 3\times1)\ \text{g/mol}=17\ \text{g/mol}$

Step2: Calculate the number of moles of $NH_3$

Using the formula $n=\frac{m}{M}$, where $m = 1.26\times 10^{4}\ \text{g}$ and $M = 17\ \text{g/mol}$. So $n=\frac{1.26\times 10^{4}\ \text{g}}{17\ \text{g/mol}}\approx741.18\ \text{mol}$

Step3: Relate moles of $NH_3$ to heat change

From the equation $N_2(g)+3H_2(g)\to2NH_3(g)\ \Delta H^{\circ}=- 92.6\ \text{kJ/mol}$ (per 2 moles of $NH_3$ produced). For $n = 741.18\ \text{mol}$ of $NH_3$, the heat change $q$ is calculated as follows:
Since for 2 moles of $NH_3$, $\Delta H=-92.6\ \text{kJ}$, for $n$ moles of $NH_3$, $q=\frac{-92.6\ \text{kJ}}{2\ \text{mol}}\times n$
Substitute $n = 741.18\ \text{mol}$ into the formula: $q=\frac{-92.6\ \text{kJ}}{2\ \text{mol}}\times741.18\ \text{mol}\approx - 3.4\times10^{4}\ \text{kJ}$ (the negative sign indicates heat is given off)

Answer:

A. $N_2(g)$

Question 6