17. determine the amount of heat, in kj, given off when 1.26 × 10⁴ g of ammonia are produced according to…

17. determine the amount of heat, in kj, given off when 1.26 × 10⁴ g of ammonia are produced according to the equation below. (molar mass nh₃ = 17.0 g/mol) n₂(g) + 3 h₂(g) → 2 nh₃(g) δh° = -92.6 kj/mol a. 3.43 × 10⁴ kj b. 6.86 × 10⁴ kj c. 7.93 × 10⁵ kj d. 1.32 × 10⁴ kj e. 2.55 × 10⁵ kj

17. determine the amount of heat, in kj, given off when 1.26 × 10⁴ g of ammonia are produced according to the equation below. (molar mass nh₃ = 17.0 g/mol) n₂(g) + 3 h₂(g) → 2 nh₃(g) δh° = -92.6 kj/mol a. 3.43 × 10⁴ kj b. 6.86 × 10⁴ kj c. 7.93 × 10⁵ kj d. 1.32 × 10⁴ kj e. 2.55 × 10⁵ kj

Answer

Explanation:

Step1: Calculate the number of moles of (NH_3)

Use the formula (n=\frac{m}{M}), where (m = 1.26\times10^{4}\space g) and (M = 17.0\space g/mol). (n=\frac{1.26\times 10^{4}\space g}{17.0\space g/mol})

Step2: Relate moles of (NH_3) to heat change

From the equation (N_{2}(g)+3H_{2}(g)\to2NH_{3}(g)\space\Delta H^{\circ}=- 92.6\space kJ/mol), for (2\space mol) of (NH_3) produced, the heat change is (-92.6\space kJ). Let (x) be the heat change for (n) moles of (NH_3). Then (x=\frac{-92.6\space kJ}{2\space mol}\times n)

Substitute (n=\frac{1.26\times 10^{4}}{17.0}) into the heat - change formula:

[ \begin{align*} x&=\frac{-92.6}{2}\times\frac{1.26\times 10^{4}}{17.0}\ &=- 46.3\times\frac{1.26\times 10^{4}}{17.0}\ &=- 46.3\times741.176\ &=-3.43\times 10^{4}\space kJ \end{align*} ]

The negative sign indicates heat is given off.

Answer:

A. (3.43\times 10^{4}\space kJ)