Question

1. Ammonia is produced directly from nitrogen and hydrogen. The chemical reaction is
N2(g) + 3 H2(g) 2 NH3(g)
N ≡ N
a) Use Table 8.4 p 326 in textbook, to estimate the enthalpy change using the bond enthalpy for the reaction. Is it exothermic or endothermic?
b) Calculate the standard enthalpy change for this reaction, using standard enthalpies of
formation ΔHf°.
2. A 1.800 g sample of phenol (C6H5OH) was burned in a bomb calorimeter whose total heat
capacity is 11.66 kJ/℃. The temp of the calorimeter plus contents increased from
22.45 to 28.27℃.
a) Write a balanced chemical equation for the bomb calorimeter reaction.
b) What is the heat of combustion/gram of phenol?
c) What is the heat of combustion/mole of phenol?

Answer 1

Answer on Question #55536 - Chemistry - General chemistry

Question:

1. Ammonia is produced directly from nitrogen and hydrogen. The chemical reaction is $\mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) 2\mathrm{NH}_3(\mathrm{g})$

N ≡ N

a) Use Table 8.4 p 326 in textbook, to estimate the enthalpy change using the bond enthalpy for the reaction. Is it exothermic or endothermic?

b) Calculate the standard enthalpy change for this reaction, using standard enthalpies of formation $\Delta Hf^{\circ}$.

2. A 1.800 g sample of phenol (C₆H₅OH) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/°C. The temp of the calorimeter plus contents increased from 22.45 to 28.27°C.

a) Write a balanced chemical equation for the bomb calorimeter reaction.

b) What is the heat of combustion/gram of phenol?

c) What is the heat of combustion/mole of phenol?

Answer:

1)

a) $\mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightarrow 2\mathrm{NH}_3(\mathrm{g})$ is exothermic reaction because $\Delta H = -91.8\ \mathrm{kJ/mol}\ (\Delta H < 0)$

b)

$$\Delta H_{f,N2} = 0\ \mathrm{kJ/mol}$$ $$\Delta H_{f,H2} = 0\ \mathrm{kJ/mol}$$ $$\Delta H_{f,NH3} = -45.9\ \mathrm{kJ/mol}$$ $$\Delta H_{\text{react}} = (2\Delta H_{f,NH3}) - (\Delta H_{f,N2} + 3\Delta H_{f,H2})$$ $$\Delta H_{\text{react}} = 2 \times (-45.9) - 0 - 3 \times 0 = -91.8\ \mathrm{kJ/mol}$$

2)

a)

$$\mathrm{C}_6\mathrm{H}_5\mathrm{OH} + 7\mathrm{O}_2 \rightarrow 6\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O}$$

b)

$$c = 11.6\ \mathrm{kJ/^\circ C}$$ $$m = 1.800\ \mathrm{g}$$ $$\Delta t = t_2 - t_1 = 28.27 - 22.45 = 5.82\ ^{\circ}\mathrm{C}$$ $$Q = C \times \Delta t$$ $$Q = 11.6 \times 5.82 = 67.86 \text{ kJ}$$ $$Q_m = Q/m = 67.86 / 1.800 = 49.18 \text{ kJ/mol}$$

c)

$$m = 1.800 \text{ g}$$ $$M = 91.4 \text{ g/mol}$$ $$n = m / M$$ $$Q_n = Q_m \times M = 49.18 \times 91.4 = 4494.5 \text{ kJ/mol}$$

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