Question

question 4 (1 point)
how much heat is 1 evolved/absorbed when 2.22 x 10² moles of br₂ are consumed? (answer = 3.26 x 10³ kj)
n₂ + 3 br₂ → 2 nbr₃ + 44.0 kj
2 kj
(2.22 x 10² mol br₂)(-------------) = 3 kj
4 mol br₂
a. 1 b. 2 c. 3 d. 4 e. 5 f. 253.719 g. 3.38 x 10⁰
h. 3.26 x 10³ i. 44.0 j. 1.33 x 10⁻² k. 9.77 x 10³ l. 4.44 x 10²
m. 6.66 x 10² n. 500.0 o. 556 p. 90.0 q. 3.2852 r. 159.81↑
s. 28.013 t. 1.750 u. 0.8750 v. 13.0 w. 0.46306 x. 1.3892
y. 24.5 z. 38.5 aa. 2.190 bb. evolved cc. absorbed
dd. exothermic ee. endothermic

Explanation:

Step1: Determine heat change type

In the reaction \(N_2 + 3Br_2
ightarrow2NBr_3+44.0\ kJ\), heat is on the product side. When \(Br_2\) is consumed (reactant side), heat is evolved (exothermic). So for blank 1, it's "evolved" (BB).

Step2: Identify heat value per mole ratio

From the reaction \(N_2 + 3Br_2
ightarrow2NBr_3+44.0\ kJ\), the ratio is \(\frac{44.0\ kJ}{3\ mol\ Br_2}\). So for blank 2, it's \(44.0\) (I), and for blank 4, it's \(3\) (A).

Step3: Calculate heat evolved

Using the formula \((2.22\times 10^{2}\ mol\ Br_2)\times(\frac{44.0\ kJ}{3\ mol\ Br_2})\).

$$
LATEXBLOCK0
$$

So for blank 3, it's \(3.26\times 10^{3}\) (H).

Answer:

1. BB. evolved
2. I. \(44.0\)
3. H. \(3.26\times 10^{3}\)
4. A. \(1\)