An equilibrium mixture of N2, H2, and NH3 at 700 K contains 0.036 M N2 and 0.15 M H2. At this temperature, Kc for the reaction N2(g) + 3 H2(g)<=> 2NH3(g) is 0.29. What is the concentration of NH3?

5.94×10¯³

Explanation:

The following data were obtained from the question:

Concentration of N2, [N2] = 0.036 M

Concentration of H2, [H2] = 0.15 M

Equilibrium constant (Kc) = 0.29 M

Concentration of NH3, [NH3] =.…?

The equation for the reaction is given below:

N2(g) + 3H2(g) <=> 2NH3(g)

Thus, we can determine the concentration of NH3 by using the equilibrium expression for the reaction.

This is illustrated below:

The equilibrium constant for a reaction is simply defined as the ratio of the concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Kc = [NH3]² / [N2] [H2]³

Inputting the value of Kc, [N2], and [H2] the value of [NH3]can be obtained as follow:

Kc = [NH3]² / [N2] [H2]³

0.29 = [NH3]²/ 0.036 × 0.15³

Cross multiply

[NH3]² = 0.29 × 0.036 × 0.15³

[NH3]² = 3.5235×10¯⁵

Take the square root of both side

[NH3] = √(3.5235×10¯⁵)

[NH3] = 5.94×10¯³

Therefore, the concentration of NH3, [NH3] is 5.94×10¯³ M.

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles

b is the correct