General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 15, Problem 15.76QP
Interpretation Introduction

Interpretation:

Calculate the partial pressure values all species of reactant and product given the equilibrium reaction at 372οC.

Concept Introduction:

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant. If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Le Chatelier's Principle (Kp): The closed system is an increase in pressure, the equilibrium will shift towards the sides of the reaction with some moles of gas. The decrease in pressure the equilibrium will shift towards the side of the reaction with high moles of gas.

Kp and Kc: This equilibrium constants of gaseous mixtures, these difference between the two constants is that Kc is defined by molar concentrations, whereas Kp is defined by the partial pressures of the gasses inside a closed system.

Expert Solution & Answer
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Answer to Problem 15.76QP

The reactant and product partial equilibrium pressure (Kp) values for the given NH3 formation reaction is shown below.

  N2(g)+3H2(g)2NH3(g)Kp=(PNH3)2(PH2)3(PN2)PN2=0.860atm; PH2=0.366atm;PNH3=4.40×10-3atm

Explanation of Solution

To find: The equilibrium reaction should be identified given the statement.

Analyze the chemical equilibrium reaction.

  N2(g)+3H2(g)2NH3(g)[FormationReaction]2NH3(g)N2(g)+3H2(g)---------[Decomposition Reaction]

The given equilibrium concentration reaction is the combined reaction is the product of the constants for this component reaction. This equilibrium reaction expression contains different conditions like solid phase into gases phase, so this process homogenous equilibrium the equilibrium constant can also be represented by Kp, were the Kp represents partial pressure. Then the each (reactant and product) molecule partial pressure (N2, H2and NH3) is derived in next method.

To find: Calculate the each partial pressure (Kp) values for given the statement of equilibrium reaction.

Calculate and analyze the (Kp) values at 3750C.

Consider the given equilibrium reactions

  N2(g)+3H2(g)2NH3(g)Initial (atm): 0.8620.3730Change (atm):  -x3x+2xEqilibrium (atm):(0.862x)(0.3733x)2xThepartialpressureequationhasKp=(PNH3)2(PH2)3(PN2)[1]Given equilibrium constan values(Kp) 4.31×10-4,PN2=0.862atm; PH2=0.373atmThe respactive values are substituted equation (1)4.31×10-4=(2x)2(0.3733x)3(0.862x)Here (3x)is very small when compared to the (0.373H2) values and that x is very small when comapred to 0.862 of (N2)values0.373-3x0.373and0.862-x0.862This values are substituted equation (1)4.31×10-4(2x)2(0.373)3(0.862)We solved here (x)4.31×10-44x2(0.0518)(0.862)=4x20.04465Changeing above equation4x2=4.31×10-4×(0.04465)4x2=1.928×105x2=1.928×1054=4.82×106Taking for square root both sidesx=2.195×10-3(Roundingtwosiginificantvalues)x=2.2×10-3The equilibrium pressure arePN2=[0.862-(2.2×10-3)]atm=0.860atmPH2=[0.373-(3)(2.2×10-3)]atm=0.366atm PNH3=(2)(2.2×10-3)atm=4.40×10-3atm 

The given ammonia formation reaction the respective reactant to give products all exists in the same phase  and this equilibrium reaction expression contains single conditions like gases phase, the equilibrium constant can also be represented by Kp, were the “P” partial pressure. The each partial pressure values are Kp derived given equation at 3750C as showed above.

Conclusion

The each of reactant and product partial pressure (Kp) values are derived given the ammonia (NH3) equilibrium reactions.

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Chapter 15 Solutions

General Chemistry

Ch. 15.4 - Practice Exercise At 430°C. the equilibrium...Ch. 15.4 - Prob. 2PECh. 15.4 - Prob. 1RCCh. 15.4 - Prob. 2RCCh. 15.4 - Prob. 3PECh. 15 - Prob. 15.1QPCh. 15 - 15.2 Explain the difference between physical...Ch. 15 - 15.3 Briefly describe the importance of...Ch. 15 - 15.4 Consider the equilibrium system 3A ⇌ B....Ch. 15 - 15.5 Define homogeneous equilibrium and...Ch. 15 - Prob. 15.6QPCh. 15 - 15.7 Write equilibrium constant expressions for...Ch. 15 - 15.8 Write the expressions for the equilibrium...Ch. 15 - 15.9 Write the equilibrium constant expressions...Ch. 15 - 15.10 Write the equation relating Kc and KP and...Ch. 15 - Prob. 15.11QPCh. 15 - Prob. 15.12QPCh. 15 - 15.13 The equilibrium constant (Kc) for the...Ch. 15 - Prob. 15.14QPCh. 15 - 15.15 What is the KP at 1273°C for the...Ch. 15 - 15.16 The equilibrium constant KP for the...Ch. 15 - Prob. 15.17QPCh. 15 - Prob. 15.18QPCh. 15 - Prob. 15.19QPCh. 15 - Prob. 15.20QPCh. 15 - Prob. 15.21QPCh. 15 - 15.22 Ammonium carbamate, NH4CO2NH2, decomposes...Ch. 15 - Prob. 15.23QPCh. 15 - 15.24 Pure phosgene gas (COCl2), 3.00 × 10−2...Ch. 15 - Prob. 15.25QPCh. 15 - 15.26 A 2.50-mol quantity of NOCl was initially...Ch. 15 - 15.27 Define reaction quotient. How does it...Ch. 15 - Prob. 15.28QPCh. 15 - Prob. 15.29QPCh. 15 - Prob. 15.30QPCh. 15 - Prob. 15.31QPCh. 15 - Prob. 15.32QPCh. 15 - Prob. 15.33QPCh. 15 - Prob. 15.34QPCh. 15 - Prob. 15.35QPCh. 15 - Prob. 15.36QPCh. 15 - Prob. 15.37QPCh. 15 - Prob. 15.38QPCh. 15 - Prob. 15.39QPCh. 15 - Prob. 15.40QPCh. 15 - Prob. 15.41QPCh. 15 - Prob. 15.42QPCh. 15 - Prob. 15.43QPCh. 15 - Prob. 15.44QPCh. 15 - Prob. 15.45QPCh. 15 - 15.46 What effect does an increase in pressure...Ch. 15 - Prob. 15.47QPCh. 15 - Prob. 15.48QPCh. 15 - 15.49 Consider the reaction Comment on the...Ch. 15 - Prob. 15.50QPCh. 15 - Prob. 15.51QPCh. 15 - Prob. 15.53QPCh. 15 - Prob. 15.54QPCh. 15 - Prob. 15.55QPCh. 15 - Prob. 15.56QPCh. 15 - Prob. 15.57QPCh. 15 - 15.58 Baking soda (sodium bicarbonate) undergoes...Ch. 15 - 15.59 Consider the following reaction at...Ch. 15 - Prob. 15.60QPCh. 15 - Prob. 15.61QPCh. 15 - Prob. 15.62QPCh. 15 - Prob. 15.64QPCh. 15 - Prob. 15.65QPCh. 15 - Prob. 15.66QPCh. 15 - Prob. 15.67QPCh. 15 - Prob. 15.68QPCh. 15 - Prob. 15.69QPCh. 15 - Prob. 15.70QPCh. 15 - Prob. 15.71QPCh. 15 - Prob. 15.72QPCh. 15 - Prob. 15.73QPCh. 15 - Prob. 15.74QPCh. 15 - Prob. 15.75QPCh. 15 - Prob. 15.76QPCh. 15 - Prob. 15.78QPCh. 15 - Prob. 15.79QPCh. 15 - Prob. 15.81QPCh. 15 - Prob. 15.82QPCh. 15 - Prob. 15.83QPCh. 15 - Prob. 15.84QPCh. 15 - Prob. 15.85QPCh. 15 - Prob. 15.86QPCh. 15 - Prob. 15.89QPCh. 15 - Prob. 15.90QPCh. 15 - Prob. 15.91QPCh. 15 - Prob. 15.92QPCh. 15 - Prob. 15.93QPCh. 15 - 15.94 Consider the decomposition of ammonium...Ch. 15 - Prob. 15.95QPCh. 15 - 15.96 In 1899 the German chemist Ludwig Mond...Ch. 15 - Prob. 15.98QPCh. 15 - Prob. 15.99QPCh. 15 - Prob. 15.100QPCh. 15 - Prob. 15.101QPCh. 15 - Prob. 15.102QPCh. 15 - Prob. 15.103SPCh. 15 - Prob. 15.104SPCh. 15 - Prob. 15.105SPCh. 15 - Prob. 15.106SPCh. 15 - Prob. 15.107SPCh. 15 - Prob. 15.110SPCh. 15 - Prob. 15.111SPCh. 15 - Prob. 15.112SP
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