Advertisement Remove all ads

Answer the following question. When 6.0 g of O2 reacts with CIF as per 2ClF(g) + O2(g) → Cl2O(g) + OF2(g) The enthalpy change is 38.55 kJ. What is standard enthalpy of the reaction? (Δr H° = 205.6 kJ) - Chemistry

Sum

Answer the following question.

When 6.0 g of O2 reacts with CIF as per 

2ClF(g) + O2(g) → Cl2O(g) + OF2(g)

The enthalpy change is 38.55 kJ. What is standard enthalpy of the reaction? (Δr H° = 205.6 kJ)

Advertisement Remove all ads

Solution

Given: 

Enthalpy change for a given mass = 38.55 kJ
Mass of O2 = 6.0 g

To find: Standard enthalpy of the given reaction

Calculation:

Number of moles of O2 = `("Mass of O"_2)/("Molar mass of O"_2) = (6 "g")/(32 "g mol"^-1)` = 0.1875 mol

The enthalpy change when 0.1875 mol of O2 react with ClF is 28.55 kJ.

∴ Enthalpy change for 1 mole O2 = `38.55/0.1875 = 205.6` kJ

From the reaction, 2 moles of ClF react with 1 mole of O2.

So, the standard enthalpy of reaction is + 205.6 kJ.

Concept: Enthalpy (H)
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 4 Chemical Thermodynamics
Exercise | Q 4.15 | Page 89
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×