Nitrosyl chloride, , decomposes to and at high temperatures, as follows.

2NOCL(g) ⇌ 2NO(g)+ Cl(g)

Suppose you place 2.00 mol NOCI in a 1.0-L flask and raise the temperature to 462°C. When equilibrium has been established, 0.66 mol NO is present. Calculate the equilibrium constant, Kc, for the decomposition reaction from these data.

Kc → 0.08

Explanation:

We propose the situations for the equilibrium:

             2NOCl (g) ⇄  2NO(g) + Cl₂(g)

Initial       2 mol                  -           -

React         x                      x           x/2

x amount has reacted before the equilibrium is reached. We got x amount of NO and x/2 of chlorine, according to stoichiometry.

Eq       2 mol - x              0.66        x/2

As we got [NO] in the equilibrum, we can determine the amount that has reacted.

0 + x = 0.66 → So 0.66 are the moles that've reacted.

2 mol - 0.66 = 1.34 moles → The amount that we had of NOCl when the requilibrium is reached.

0.66 /2 = 0.33 moles of Cl₂

Let's make the expression for Kc

Kc = [NO]² . [Cl₂] / [NOCl]²

Kc = 0.66² . 0.33 / 1.34²

Kc = 0.08

The equilibrium constant Kc = 0.080

Explanation:

Step 1: data given

Number of moles NOCl = 2.00 moles

Volume = 1.00 L

After heating the number of moles NO = 0.66 moles

Step 2: The balanced equation

2NOCL(g) ⇌ 2NO(g)+ Cl2(g)

Step 3: Calculate initial concentration

Concentration = moles / volume

[NOCl] = 2.00 moles / 1L = 2 M

[NO] = 0M

[Cl2]= 0M

Step 4: Calculate concentration at the equilibrium

[NOCl] = 2M - 2X

[NO] = 2X M = 0.66 M

[Cl2]= XM

X = 0.66/2 = 0.33

[NOCl] = 2M - 2*0.33 = 1.34 M

[NO] = 2X M = 0.66 M

[Cl2]= XM = 0.33 M

Step 5: Calculate Kc

Kc = [Cl2]*[NO]² / [NOCl]²

Kc = (0.33* 0.66² )/ 1.34²

Kc = 0.080

The equilibrium constant Kc = 0.080

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol