Question

pter 4-6 0 Saved Help Save At 25°C, the following heats of reaction are known: 2C1F(g) + O2(g) → Cl2O(g) +F20(g) 2C1F3(g) + 2O2(g) + Cl2O(g) + 3F2O(g) 2F2(g) + O2(g) → 2F20(g) AHO/= 167.4 kJ/mol AHºrn = 341.4 kJ/mol AHºrn=-43.4 kJ/mol At the same temperature, use the above data to calculate the heat released (kJ) when 3.40 moles of CIF(g) reacts with excess F2. CIF(g) + F2(9) --> CIF3(9) Write answer to three significant figures. NO SIGN in ANSWER. Numeric Response < Prey 22 of 25 Next > arch

Answer 1

Given data,

2 ClF (g) + O2 (g) ------> Cl2O (g) + F2O (g) $\Delta$ H1 = 167.4 kJ/mol ----------- (1)

2 ClF3 (g) + 2 O2 (g) ----------> Cl2O (g) + 3 F2O (g) ,  $\Delta$H2 = 341.4 kJ/mol ---------- (2)

2 F2 (g) + O2 (g) ----> 2 F2O (g) $\Delta$ H3 = - 43.4 kJ/mol ----------- (3)

Required equation is,

ClF (g) + F2 (g) ------------> ClF3 (g) $\Delta$ H = ?

To get the required equation, first, do (1) + (3) - (2)

2 ClF (g) + O2 (g) ------> Cl2O (g) + F2O (g) $\Delta$ H1 = 167.4 kJ/mol ----------- (1)

(+)

2 F2 (g) + O2 (g) ----> 2 F2O (g) $\Delta$ H3 = - 43.4 kJ/mol ----------- (3)

( - )

2 ClF3 (g) + 2 O2 (g) ----------> Cl2O (g) + 3 F2O (g) ,  $\Delta$H2 = 341.4 kJ/mol ---------- (2)

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2 ClF (g) + 2 F2 (g) ------------> 2 ClF3 (g) $\Delta$ H4 = ?

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Therefore,

$\Delta$H4 = $\Delta$ H1 + $\Delta$ H3 - $\Delta$ H2

$\Delta$H4 = 167.4 + ( - 43.4 ) - 341.4

$\Delta$H4 = - 217.4

This is the heat released when 2 mol of ClF reacts with excess F2.

Then, heat released when 3.40 moles of ClF reacts with excess F2 is,

$\Delta$H =

-XΔΗ4

$\Delta$H =

3.4 x(–217.4)

$\Delta$H = - 370 kJ