. Consider the reaction: H2(g) + I2(g) ? 2HI(g) A reaction mixture at equilibrium at 175...
. Consider the reaction: H2(g) + I2(g) ? 2HI(g)
A reaction mixture at equilibrium at 175 K contains PH2
= 0.958 , PI2
= 0.877 ,
and PHI= 0.020 . A second
mixture not equilibrium, but at the same temperature,contains
PH2 = PI2 = 0.621 , and PHI = 0.101
. a) For the second mixture, calculate Q.
b) Determine whether the reaction in the second mixture will shift right or shift left.c) Construct an ICE table. Keep in mind your answer for a in regards to the sign of the
Homework Answers
H2(g) + I2(g) ? 2HI (g)
0.958atm 0.020atm 0.020atm Mixture 1
0.621atm 0.621atm 0.101atm Mixture 2
Kp1 = (0.020)2
(0.958)(0.877)
= 4.76
Add Answer to:
. Consider the reaction: H2(g) + I2(g) ? 2HI(g)
A reaction mixture at equilibrium at 175...
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a) H2(g) + 12 (g) = 2 HI(g) (3 points) A reaction mixture at equilibrium at 175 K contains Pue = 0.958 am, Pe=0.877 atm, and Pu=0.020 atm. A second reaction mixture, also at 175 K. contains Pie Pe=0.621 atm, and Pm 0.101 atm. Is the second reaction mixture at equilibrium? (Show all work for full credit) a IT b) (7 points) If not at equilibrium, what will be the partial pressure of HI when the reaction reaches equilibrium at...
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1) At 298 K, an equilibrium mixture contains P H2=0.958 atm, P I2= 0.877 atm and...
1) At 298 K, an equilibrium
mixture contains P H2=0.958 atm, P I2= 0.877 atm and P HI=0.020
atm. Please try and answer all!
You must show all your work to receive credit. Watch sig figs! Consider the reaction: 1) At 298 K, an equilibrium mixture contains P H2-0.958 atm, P ½=0.877 atm, and P HI-0.020 atm. Calculate the value of the equilibrium constant Kp? a) (o, o2) 0.938x o.87? b) And what is the value of the equilibrium constant...
For the reaction H2 (g) + I2 (g) = 2HI (g); Kc =50.0. Calculate the concentration...
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Consider this reaction at 298 K: H2 (g) + I2 (g) ⇌ 2 HI(g) Calculate ΔGrxn under the following conditions: PH2 (g) = 0.161 atm PI2 (g) = 0.186 atm PHI(g) = 0.307 atm
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The equilibrium constant for the following reaction is 1.80x102 at 698K 2HI(8)H2( ) +I2(g) If an equilibrium mixture of the three gases at 698K contains 2.35x102 M HI(g) and 2.63x10-2 M H, what is the equilibrium concentration of I? Submit Answer Retry Entire Group 9 more group attempts remaining
a) H2(g) + 12 (g) = 2 HI(g) (3 points) A reaction mixture at equilibrium at 175 K contains Pue = 0.958 am, Pe=0.877 atm, and Pu=0.020 atm. A second reaction mixture, also at 175 K. contains Pie Pe=0.621 atm, and Pm 0.101 atm. Is the second reaction mixture at equilibrium? (Show all work for full credit) a IT b) (7 points) If not at equilibrium, what will be the partial pressure of HI when the reaction reaches equilibrium at...
Given the equilibrium reaction: 2HI(g) H2(g) + I2(g) A sample mixture of HI, H2, and 12, at equilibrium, was found to have [H2]- 1.4 x 102 Mand [HI 4.0 x 102 M. If Keq 1.0 x 10, calculate the molar concentration of I2 in the equilibrium mixture, Enter your answer in the provided box. ]= м
1) At 298 K, an equilibrium
mixture contains P H2=0.958 atm, P I2= 0.877 atm and P HI=0.020
atm. Please try and answer all!
You must show all your work to receive credit. Watch sig figs! Consider the reaction: 1) At 298 K, an equilibrium mixture contains P H2-0.958 atm, P ½=0.877 atm, and P HI-0.020 atm. Calculate the value of the equilibrium constant Kp? a) (o, o2) 0.938x o.87? b) And what is the value of the equilibrium constant...
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