The value of AH for the reaction below is-72 kJ.kJ of heat are released when 1.0

Question

The value of AH for the reaction below is-72 kJ.kJ of heat are released when 1.0 mol of HBr is formed in this reaction. H2 (g) + Br2 (g) 2HBr (g) 0.44 -72 , 144 36 72 3.34 p QUESTION 16 Which of the following is/are an endothermic process? Both boiling soup and ice melting Hydrochloric acid and barium hydroxide are mixed at 25°C: the temperature increases water freezing boiling soup ice melting 3.34 p QUESTION 17 Automobile air bags use the decomposition of sodium azide as their source of gas for rapid inflation 2NaN3 (s) 2Na (s) + 3N2 (g). What mass (g) of NaN3 is required to provide 40.0 L of N2 at 25.0'C and 763 torr? 71.1 160 1.09 107

Explanation / Answer

When 2 moles of HBr are formed, 72 KJ heat is released

When 1 mole of HBr is formed 72/2 KJ = 36 KJ heat is released

Released heat = 36 KJ

QUESTION NO. 16

Endothermic process is-

Both boiling soup and melting ice

QUESTION NO. 17

Let us first find the no. of moles of N2 produced. P = 763 torr = (763/760) atm = 1.004 atm, V= volume = 40 litres, T = temperature = (273+25)K = 298 K

From, ideal gas law we get, P * V = n * R *T

No. of moles of N2 = n = (P* V)/ (R * T) = (1.004 atm * 40 Lit)/ (0.082 Lit-atm-Mol-1-K-1 * 298 K) = 1.64 mole

Look at the stoichiometry of the reaction:

3 moles N2 is generated from 2 moles of NaN3

1.64 moles of N2 is generated from (2*1.64/3) moles = 1.093 mole of NaN3

Molecular weight of NaN3 is = 65 g/ mole

So, 1.093 mole NaN3 = 1.093 * 65 grams = 71.1 grams NaN3

Answer is 71.1 grams

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