When 1 mol of KBr(s) decomposes to its elements, 394 kJ of heat is absorbed. (b) What is ΔH when 10.0 kg of KBr forms from its elements?

Answer :

The HEAT released when 10.0 kg of KBr forms from its element is -3. 31 × 10^4 kJ

As we know that,

thermochemical equation for the decomposition of potassium bromide is

KBr ------- K + 1/2 Br2

Now, reversing the above equation the sign of heat also become opposite. Thus, reverse reaction can be written as

K + 1/2 Br2 ----KBr

∆H = - 394 kJ.

Given,

Molar mass of KBr = 119 g/mol

1 mol KBr = -394kJ

As we know that

1 kg = 10^3

Calculation of conversation factor

The desired conversation factor can be calculated as

10^3 × (1mol KBr/ 119 g KBr) × (-394KJ/ 1 mol KBr)

Calculation of heat released

heat released = given mass × conversion factor

= 10 kg KBr × 10^3 × (1mol KBr/ 119 g KBr) × (-394KJ/ 1 mol KBr)

= -3. 31 × 10^4 kJ

Thus, we calculated that the heat released when 10.0 kg of KBr forms from its element is -3. 31 × 10^4 kJ.

learn more about decomposition reaction:

https://brainly.com/question/2733624

#SPJ4

Other Questions