I need help with part A. Problem 15.35 A mixture of 0.10 mol of NO, 0.050 mol of

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I need help with part A.

Problem 15.35

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established:
2NO(g)+2H2(g)N2(g)+2H2O(g)
At equilibrium [NO]=0.062M.

Part A

Calculate the equilibrium concentration of H2.

Express your answer using two significant figures.

.062

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Incorrect; Try Again

Part B

Calculate the equilibrium concentration of N2.

Express your answer using two significant figures.

1.9×102

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Correct

Part C

Calculate the equilibrium concentration of H2O.

0.138

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Correct

Part D

Calculate Kc.

Express your answer using two significant figures.

650

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Correct

Problem 15.35

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established:
2NO(g)+2H2(g)N2(g)+2H2O(g)
At equilibrium [NO]=0.062M.

Part A

Calculate the equilibrium concentration of H2.

Express your answer using two significant figures.

.062

M

SubmitMy AnswersGive Up

Incorrect; Try Again

Part B

Calculate the equilibrium concentration of N2.

Express your answer using two significant figures.

1.9×102

M

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Correct

Part C

Calculate the equilibrium concentration of H2O.

0.138

M

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Correct

Part D

Calculate Kc.

Express your answer using two significant figures.

Kc =

650

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Correct

Explanation / Answer

2NO(g)+2H2(g)N2(g)+2H2O(g)

[NO] = 0.1

[H2] = 0.05

[N2] = 0

[H2O] = 0.1

in equilibrium

[NO] = 0.1 -2x

[H2] = 0.05 -2x

[N2] = 0 +x

[H2O] = 0.1 +2x

and [NO] = 0.1 -2x = 0.062

x = (0.062-0.1)/(-2) = 0.019

[NO] = 0.1 -2x = 0.1-2*0.019 = 0.062

[H2] = 0.05 -2x = 0.05-2*0.019 = 0.012

[N2] = 0 +x = 0.019

[H2O] = 0.1 +2x = 0.1 +2*0.019 = 0.138

then

K = [N2][H2O]^2/[NO]^2[H2]^2

K = (0.019)(0.138^)/((0.062^2)(0.012^2))

Kc = 653.681

if needed to 2 sig fig... 653.681 --> 650

which is pretty proximate to 650

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