Hydrogen gas at a temperature of 22.0°C that is confined in a 5.00L cylinder exerts a pressure of 4.20atm. If the gas is released into a 10.0L reaction vessel at a temperature of 33.6°C, what will be the pressure inside the reaction vessel?

The pressure inside the reaction vessel is 2.18 atm

Explanation:

Step 1: Data given

The temperature = 22.0 °C = 295.15 K

Pressure = 4.20 atm

Volume = 5.0 L

Volume increased to 10.0 L

Temperature increased to 33.6 °C = 306.75K

Step 2: Calculate the new pressure

P1 * V1 / T1 = P2*V2 / T2

⇒with P1 = the initial pressure = 4.20 atm

⇒with V1 = the initial volume = 5.0 L

⇒with T1 = the initial temperature = 295.15 K

⇒with P2 = the new pressure = TO BE DETERMINED

⇒with V2 = the increased volume = 10.0L

⇒with T2 = the increased temperature = 306.75 K

(4.20 atm * 5.0 L) / 295.15 K = ( P2 * 10.0 L ) / 306.75 K

P2 = (4.20 * 5.0 L * 306.75 K) / (295.15 K * 10.0 L)

P2 = 2.18 atm

The pressure inside the reaction vessel is 2.18 atm

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :