Holooly Plus Logo
Holooly Plus Logo

Question 7.10: Le Châtelier’s Principle and Equilibrium Mixtures Nitrogen r......

Le Châtelier’s Principle and Equilibrium Mixtures
Nitrogen reacts with oxygen to give NO:

N_2(g) + O_2(g) \rightleftarrows  2 NO(g)    ΔH = +43 kcal/mol (+180 kJ/mol)

Explain the effects of the following changes on reactant and product concentrations:
(a) Increasing temperature                          (b) Increasing the concentration of NO
(c) Adding a catalyst

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.
Report Answer

(a) The reaction is endothermic (positive ΔH), so increasing the temperature favors the forward reaction. The concentration of NO will be higher at equilibrium.
(b) Increasing the concentration of NO, a product, favors the reverse reaction. At equilibrium, the concentrations of both N_2 and O_2, as well as that of NO, will be higher.
(c) A catalyst accelerates the rate at which equilibrium is reached, but the concentrations at equilibrium do not change.

Related Answered Questions

Question: 7.1

Heat of Reaction from Bond Energies Estimate the ΔH (in kcal/mol) for the reaction of hydrogen and oxygen to form water: 2H2 + O2 → 2H2O ΔH = ? ANALYSIS The individual bond energies from Table 7.1 can be used to calculate the total bond energies of reactants and products. ΔH can then be calculated ...

Verified Answer:

ΔH = ∑(Bond dissociation energies)_{reactan...
Question: 7.9

Equilibrium Equations: Calculating K In the reaction of Cl2 with PCl3, the concentrations of reactants and products were determined experimentally at equilibrium and found to be 7.2 mol/L for PCl3, 7.2 mol/L for Cl2, and 0.050 mol/L for PCl5. PCl3(g) + Cl2(g) ↔ PCl5(g) Write the equilibrium ...

Verified Answer:

K = \frac{[PCl_5 ]}{[PCl_3 ] [Cl_2 ]} = \fr...
Question: 7.2

Heat of Reaction: Moles Methane undergoes combustion with O2 according to the following equation: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔH = -213 kcal/mol CH4 = -891 kJ/mol CH4 How much heat (in kcal and kJ) is released during the combustion of 0.35 mol of methane? ANALYSIS Since the value of ΔH ...

Verified Answer:

To find the amount of heat released (in kilocalori...
Question: 7.3

Heat of Reaction: Mass to Mole Conversion How much heat is released during the combustion of 7.50 g of methane (molar mass = 16.0 g/mol)? CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔH = -213 kcal/mol CH4 = -891 kJ/mol CH4 ANALYSIS We can find the moles of methane involved in the reaction by using the ...

Verified Answer:

Going from a given mass of methane to the amount o...
Question: 7.4

Heat of Reaction: Mole Ratio Calculations How much heat is released in kcal and kJ when 2.50 mol of O2 reacts completely with methane? CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔH = -213 kcal/mol CH4 = -891 kJ/mol CH4 ANALYSIS Since the ΔH for the reaction is based on the combustion of 1 mol of methane, ...

Verified Answer:

To find the amount of heat released by combustion ...
Question: 7.6

Spontaneity of Reactions: Enthalpy, Entropy, and Free Energy The industrial method for synthesizing hydrogen by reaction of carbon with water has ΔH = +31.3 kcal/mol (+131 kJ/mol) and ΔS = +32 cal/ [mol • K](+134 J/ [mol • K]). What is the value of ΔG (in kcal and kJ) for the reaction at 27 °C (300 ...

Verified Answer:

Use the free-energy equation to determine the valu...
Question: 7.8

Writing Equilibrium Equations The first step in the industrial synthesis of hydrogen is the reaction of steam with methane to give carbon monoxide and hydrogen. Write the equilibrium equation for the reaction. H2O(g) + CH4(g) ↔ CO(g) + 3 H2(g) ANALYSIS The equilibrium constant K is the number ...

Verified Answer:

K = \frac{[CO] [H_2]^ 3}{[H_{2}O] [CH_4]}[/...
Question: 7.5

Entropy Change of Processes Does entropy increase or decrease in the following processes? (a) Smoke from a cigarette disperses throughout a room rather than remaining in a cloud over the smoker’s head. (b) Water boils, changing from liquid to vapor. (c) A chemical reaction occurs: 3 H2(g) + N2(g) → ...

Verified Answer:

(a) Entropy increases because smoke particles are ...
Question: 7.7

Energy of Reactions: Energy Diagrams Draw an energy diagram for a reaction that is very fast but has a small negative free-energy change. ANALYSIS A very fast reaction has a small Eact. A reaction with a small negative free-energy change is a favorable reaction with a small energy difference ...

Verified Answer: