2NH3(g) CO2(g) In an experiment carried out at this temperature, a certain amount of NH4OCONH2 is placed in an evacuated rigid container and allowed to come to equilibrium. Calculate the total pressure in the container at equilibrium.

The question is incomplete, here is the complete question:

At 25°C, Kp = 2.9 × 10⁻³ for the reaction:

In an experiment carried out at this temperature, a certain amount of NH₄OCONH₂ is placed in an evacuated rigid container and allowed to come to equilibrium. Calculate the total pressure in the container at equilibrium.

The total pressure in the container at equilibrium is 0.2694 atm

Explanation:

Let the initial concentration of be 'x'

The given chemical equation follows:

Initial:                       x

At eqllm:              x-y                         2y              y

The expression of for above equation follows:

The partial pressure of pure solids and liquids are taken as 1 in equilibrium constant expression. So, the partial pressure of is not seen in the expression.

We are given:

Putting values in above expression, we get:

So, the equilibrium partial pressure of ammonia = 2y = (2 × 0.0898) = 0.1796 atm

The equilibrium partial pressure of carbon dioxide = y = 0.0898 atm

Total pressure inside the container at equilibrium =

Hence, the total pressure in the container at equilibrium is 0.2694 atm

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

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