Select a reaction by clicking and dragging the reaction into one of the spaces b

Question

Select a reaction by clicking and dragging the reaction into one of the spaces below. Reactions delta_r H degrees (kJ/mol) SnCl_2(s) + TiBr_2 (s) rightarrow SnBr_2(s) + TiCl_2(s) +4.2 SnCl_2(s) + Cl_2(g) rightarrow SnCl_4(C) -195 TiCl_2(s) + Cl_2(g) rightarrow TiCl_4(C) -273 delta _r H degree (kJ/mol) Ckar Reverse ______________ Ckar Reverse ______________ Ckar Reverse _______________ _______________ Net Reaction: delta_rH^degree_net = 0.0 kJ/mol Select reactions to be obtain this reaction: SnCl_2(s) + TiCl_4(C) rightarrow TiCl_2(s) + SnCl_4(C) When this net reaction has been produced, press Check.

Explanation / Answer

The three given reactions are:

SnCl2 (s) + TiBr2 (s) -----> SnBr2 (s) + TiCl2 (s); H0 = +4.2 kJ/mol ….(R1)

SnCl2 (s) + Cl2 (g) -----> SnCl4 (l); H0 = -195 kJ/mol ….(R2)

TiCl2 (s) + Cl2 (g) -----> TiCl4 (l); H0 = -273 kJ/mol ….(R3)

Reverse (R3) (this will also change the sign of H0) and add to R2 to get

SnCl2 (s) + Cl2 (g) + TiCl4 (l) -----> SnCl4 (l) + TiCl2 (s) + Cl2 (g)

Cancel out common terms from both sides to get

SnCl2 (s) + TiCl4 (l) -----> TiCl2 (s) + SnCl4 (l)

This is the given reaction as you wanted. However, the reaction will not have H0 = 0 since H0rxn = H0 (R2) + (-H0) (R3) (note we have reversed R3) = (-195 kJ/mol) + [-(-273 kJ/mol)] = (-195 kJ/mol) + (273 kJ/mol) = +78 kJ/mol. Please check.

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