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Chapter 6: Problem 139

Calculate \(q, w, \Delta E,\) and \(\Delta H\) for the process in which 88.0 g of nitrous oxide (laughing gas, \(\mathrm{N}_{2} \mathrm{O} )\) is cooled from \(165^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\) at a constant pressure of 5.00 \(\mathrm{atm} .\) The molar heat capacity for $\mathrm{N}_{2} \mathrm{O}(g)\( is 38.7 \)\mathrm{J} /^{\prime} \mathrm{C} \cdot$ mol.

Short Answer

Expert verified
The heat transfer (q) for the given process is -8522.0 J, the work done (w) cannot be calculated due to insufficient information, the change in internal energy (ΔE) is -8522.0 J + w, where w is the work done, and the change in enthalpy (ΔH) is -8522.0 J.

Step by step solution

01

1. Convert mass of N2O to moles

To calculate moles of N2O, we need to use the molar mass of N2O. The molar mass of Nitrous Oxide (N2O) is: 28.0134 g/mol (N) + 2 * 15.9994 g/mol (O) = 44.0128 g/mol Given mass = 88.0 g Moles of N2O = Mass (g) / Molar Mass Moles of N2O = 88.0 g / 44.0128 g/mol = 2.0 mol
02

2. Calculate the change in temperature

ΔT = Tfinal - Tinitial ΔT = 55°C - 165°C = -110°C
03

3. Calculate the heat transfer (q)

q = n * C_p * ΔT Where n = moles of N2O, and C_p = molar heat capacity q = (2.0 mol) * (38.7 J/mol °C) * (-110 °C) q = -8522.0 J
04

4. Calculate the work done (w)

At constant pressure, the work done is given as: w = -P * ΔV Since the information on the change in volume is not provided, we can proceed to calculate ΔE without w.
05

5. Calculate the change in internal energy (ΔE)

ΔE = q + w Without the work (w) value, we can't find the exact value for ΔE. However, we can find the relationship between ΔE and the work (w). ΔE = -8522.0 J + w
06

6. Calculate the change in enthalpy (ΔH)

At constant pressure, the change in enthalpy (ΔH) can be calculated as: ΔH = q Since the heat transfer (q) is -8522.0 J, ΔH = -8522.0 J
07

Results

: q = -8522.0 J w = Not enough information ΔE = -8522.0 J + w ΔH = -8522.0 J

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Most popular questions from this chapter

Consider the following reaction: $$\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$ \Delta H=-891 \mathrm{kJ} $$ Calculate the enthalpy change for each of the following cases: a. 1.00 g methane is burned in excess oxygen. b. \(1.00 \times 10^{3}\) L methane gas at 740 . torr and $25^{\circ} \mathrm{C}$ are burned in excess oxygen.

A swimming pool, 10.0 \(\mathrm{m}\) by \(4.0 \mathrm{m},\) is filled with water to a depth of 3.0 \(\mathrm{m}\) at a temperature of \(20.2^{\circ} \mathrm{C}\) . How much energy is required to raise the temperature of the water to \(24.6^{\circ} \mathrm{C} ?\)

The bomb calorimeter in Exercise 112 is filled with 987 \(\mathrm{g}\) water. The initial temperature of the calorimeter contents is $23.32^{\circ} \mathrm{C} .\( A \)1.056-\mathrm{g}\( sample of benzoic acid \)\left(\Delta E_{\mathrm{comb}}=\right.\( \)-26.42 \mathrm{kJ} / \mathrm{g}$ ) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?

Consider the reaction $$ 2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ $$ \Delta H=-118 \mathrm{kJ} $$ Calculate the heat when 100.0 \(\mathrm{mL}\) of 0.500\(M \mathrm{HCl}\) is mixed with 300.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{Ba}(\mathrm{OH})_{2}\) . Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of 400.0 \(\mathrm{g}\) and a specific heat capacity of 4.18 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) , calculate the final temperature of the mixture.

The enthalpy change for a reaction is a state function and it is an extensive property. Explain.
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