Vithdr... Forms | Office of t... KB Viewing Your Aca....3The equilibrium constant, Ke, for the following reaction is 7.00 x 10-5 at 673 K.Submit Answer$4888NH₂I(8) NH3(g) + HI(g)If an equilibrium mixture of the three compounds in a 7.23 L container at 673 K contains 2.31 mol of NH4I(s) and 0.405mol of NH3, the number of moles of HI present ismol.RF4%5Show HintRetry Entire Group 9 more group attempts remainingT[Review Topics][References]Use the References to access important values if needed for this question.F5Cengage Learning Cengage Technical SupportA6Have Changes in..MacBook AirF6YF G H&7◄◄U-*Scholarship Ameri.* 008► 1119K►►F9O)0bio 1108 chatAF10PTpSAVAGE X FENTY...PreviousF1111 +NextSave and Exit4) F12G

we need to calculate moles of HI at equilibrium

Use the References to access important values if needed for this question. The equilibrium constant, Ke, for the following reaction is 7.00 x 10-5 at 673 K. NH₂I(8) NH3(g) + HI(g) If an equilibrium mixture of the three compounds in a 7.23 L container at 673 K contains 2.31 mol of NH4I(s) and 0.405 mol of NH3, the number of moles of HI present is mol.

Use the References to access important values if needed for this question. The equilibrium constant, Ke, for the following reaction is 7.00 x 10-5 at 673 K. NH₂I(8) NH3(g) + HI(g) If an equilibrium mixture of the three compounds in a 7.23 L container at 673 K contains 2.31 mol of NH4I(s) and 0.405 mol of NH3, the number of moles of HI present is mol.

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter12: Chemical Equilibrium

Section: Chapter Questions

Problem 12.22PAE

See similar textbooks

Similar questions

A common type of reaction we will study is that having a very small K value (K 1). Solving for equilibrium concentrations in an equilibrium problem usually requires many mathematical operations to be perfomed. However, the math involved when solving equilibrium problems for reactions having small K values (K 1) is simplified. What assumption is made when solving the equilibrium concentrations for reactions with small K values? Whenever assumptions are made, they must be checked for validity. In general, the 5% rule is used to check the validity of assuming x (or 2 x, 3x, and so on) is very small compared to some number. When x (or 2 x. 3x. and so on) is less than 5% of the number the assumption was made against, then the assumption is said to be valid. If the 5% rule fails, what do you do to solve for the equilibrium concentrations?
Again the experiment in Exercise 12.33 was redesigned. This time, 0.15 mol each of N, and O2 was injected into a 5.0-L container at 2500 K, at which the equilibrium constant is 3.6 X 10-?. What was the composition of the reaction mixture at equilibrium? l'he following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) *2 2 NO K = 4.1 X IO-4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L-’ ofO2, what were the equilibrium concentrations of all species?
Data for the reaction [Mn(CO)5(CH3CN)]+ + NC5H5 (Mn(CO)5(NC5H5)]+ + CH3CN are given in the table. Calculate Ea from a plot of In k versus 1/T.
1’he reaction in Exercise 12.33 was repeated. This time, the reaction began when only NO was injected into the reaction container. 110.200 mol L_l NO was injected, what were the equilibrium concentrations of all species? The following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) ^2 NO K = 4.1 X 10~4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L"' ofO2, what were the equilibrium concentrations of all species?
Because carbonic acid undergoes a second ionization, the student in Exercise 12.39 is concerned that the hydrogen ion concentration she calculated is not correct. She looks up the equilibrium constant for the reaction HCO,-(aq) «=* H+(aq) + COf'(aq) Upon finding that the equilibrium constant for this reaction is 4.8 X 10“H, she decides that her answer in Exercise 12.39 is correct. Explain her reasoning. A student is simulating the carbonic acid—hydrogen carbonate equilibrium in a lake: H,CO,(aq) 5=6 H+(aq) + HCO,'(aq) K = 4.4 X 10'7She starts with 0.1000 A1 carbonic acid. W hat are the concentrations of all species at equilibrium?
Consider the system A(g)+2B(g)+C(g)2D(g)at 25C. At zero time, only A, B, and C are present. The reaction reaches equilibrium 10 min after the reaction is initiated. Partial pressures of A, B, and D are written as PA, PB, and PD. Answer the questions below, using LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required). (a) PD at 11 min PD at 12 min. (b) PA at 5 min PA at 7 min. (c) K for the forward reaction K for the reverse reaction. (d) At equilibrium, K Q. (e) After the system is at equilibrium, more of gas B is added. After the system returns to equilibrium, K before the addition of B K after the addition of B. (f) The same reaction is initiated, this time with a catalyst K for the system without a catalyst the system with a catalyst. (g) K for the formation of one mole of D K for the formation of two moles of D. (h) The temperature of the system is increased to 35C. PB at equilibrium at 25C PB at equilibrium at 35C. (i) Ten more grams of C are added to the system. PB before the addition of C PB after the addition of C.
In Table 12.1 (←Sec. 12-3a) the equilibrium constant for the reaction ⇋ is given as 4.2 × 1052. If this reaction is so product-favored, why can large piles of yellow sulfur exist in our environment (as they do in Louisiana and Texas)?
The experiment in Exercise 12.33 was redesigned so that the reaction started with 0.15 mol each of N2 and O2 being injected into a 1.0-L container at 2500 K. The equilibrium constant at 2500 K is 3.6 X 10“’. What was the composition of the reaction mixture after equilibrium was attained? The following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) *2 2 NO K = 4.1 X IO-4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L-’ ofO2, what were the equilibrium concentrations of all species?
Because calcium carbonate is a sink for CO32- in a lake, the student in Exercise 12.39 decides to go a step further and examine the equilibrium between carbonate ion and CaCOj. The reaction is Ca2+(aq) + COj2_(aq) ** CaCO,(s) The equilibrium constant for this reaction is 2.1 X 10*. If the initial calcium ion concentration is 0.02 AI and the carbonate concentration is 0.03 AI, what are the equilibrium concentrations of the ions? A student is simulating the carbonic acid—hydrogen carbonate equilibrium in a lake: H2COj(aq) H+(aq) + HCO}‘(aq) K = 4.4 X 10"7 She starts with 0.1000 AI carbonic acid. What are the concentrations of all species at equilibrium?