Write the Kc expression, set up an ICE chart, substitute, and solve.
moles H2 = 0.763/2.016 and that divided by 3.67 L = M; approximately 0.1 but you need to do it more precisely than that.
moles I2 = 96.9g/molar mass I2 and that divided by 3.67 L = about 0.1 M. Again you should confirm that. It's slight more than that.
moles HI = 90.4/molar mass HI and that divided by 3.67 L for M = about 0.192. Check my numbers.
Then the ICE chart.
initial:
H2 = about 0.1 M
I2 = about 0.1 M
HI = 0
change:
H2 = -x
I2 = -x
HI = +2x
equilibrium:
H2 = 0.1-x
I2 = 0.1-x
HI = 0.192
SO, if HI started out at 0 and it end up at 0.192, then 0.192 must equal to 2x or x = 0.192/2 = 0.096 M
Then H2 at equilibrium must be0.1-0.096 and
I2 must be 0.1-0.096.
Substitute those into Kc expression. and solve for K.
Check my numbers for sure.
I stuck on this problem.
Can you help me to solve it?
A reaction mixture in a 3.67L flask at a certain temp. initially contains 0.763g H2 and 96.9g I2. At equilibrium, the flask contains 90.4g HI. Calculate the equilibrium constant(Kc) for the reaction at this temp.
H2(g) + I2(g) <-> 2HI(g)
PLZZZ help me!!
4 answers
You must use the I.C.E. table for this problem. First, you must convert the grams given into M (molarity)which is in(moles/L).
Molar Masses:
H2= 1.01x2 = 2.02g/mol
I2= 126.90x2 = 253.80g/mol
HI= 1.01+126.90=127.91g/mol
Converting to Moles:
0.763g of H2/(mol/2.02g)=0.378mol
96.9g of I2/(mol/253.80g)=0.382mol
90.4g of HI/(mol/127.91g/mol)=0.707mol
Converting to M:
H2=0.378mol/3.67L=0.103M
I2=0.382mol/3.67L=0.104M
HI=0.707mol/3.67L=0.193M
You must have M (at least for this question) to make an I.C.E. table.
I.C.E. Table:(M)
H2 I2 2HI Kc
I:0.103 0.104 0.00 0.00
C:-0.0965 -0.0965 +0.193 (an increase)
E:0.0065 0.0075 0.193 ?
Solve for Equilibrium Constant (Kc):
(0.193)^2/(0.0065)(0.0075)
Kc=764!!!
Hope this helped unlike this guy^ ;)
Molar Masses:
H2= 1.01x2 = 2.02g/mol
I2= 126.90x2 = 253.80g/mol
HI= 1.01+126.90=127.91g/mol
Converting to Moles:
0.763g of H2/(mol/2.02g)=0.378mol
96.9g of I2/(mol/253.80g)=0.382mol
90.4g of HI/(mol/127.91g/mol)=0.707mol
Converting to M:
H2=0.378mol/3.67L=0.103M
I2=0.382mol/3.67L=0.104M
HI=0.707mol/3.67L=0.193M
You must have M (at least for this question) to make an I.C.E. table.
I.C.E. Table:(M)
H2 I2 2HI Kc
I:0.103 0.104 0.00 0.00
C:-0.0965 -0.0965 +0.193 (an increase)
E:0.0065 0.0075 0.193 ?
Solve for Equilibrium Constant (Kc):
(0.193)^2/(0.0065)(0.0075)
Kc=764!!!
Hope this helped unlike this guy^ ;)
Use the excerpt Proposed Equal Rights Amendment (ERA) to answer the question.
What was the outcome of the ERA? Explain
A. It failed to be ratified by state legislature.
B. It was ratified through a congressional vote.
C. It was ratified by three-quarters of the state legislatures.
D. It failed to be ratified by three-quarters of the House and Senate.
What was the outcome of the ERA? Explain
A. It failed to be ratified by state legislature.
B. It was ratified through a congressional vote.
C. It was ratified by three-quarters of the state legislatures.
D. It failed to be ratified by three-quarters of the House and Senate.
A. It failed to be ratified by state legislatures.
3 answers