Question

Chemistry QB14.32 Level 2

х Question The reaction 2 C[math]_{2}[/math]H[math]_{6}[/math](g) + 5 O[math]_{2}[/math](g) [math]\leftrightharpoons[/math] 4 CO(g) + 6 H[math]_{2}[/math]O(g) has [math]\Delta[/math]H[math]^{0}_{rxn}[/ma<0. Determine for each indicated change whether the reactant(s) or product(s) concentrations will [math]\textbf[increase}[/math] or [math]\textbf{not change}[/math] when the reaction reestablishes equilibrium. Your answers must be [math]\textbf{reactants}[/math], [math]\textbf{products}[/math], or [math]\textbf(not change}[/math). Increase the C[math]_{2}[/math]H[math]_{6}[/math] partial pressure. Increase the O[math]_{2}/math] concentration Decrease the CO partial pressure. Add 2.0 moles of H[math]_{2}[/math]o. Decrease the reaction vessel volume. Add a catalyst to the reaction mixture. Increase the reaction temperature. 5 attempts left. Submit

Answer 1

2C2H6 + 5O2 --‐-> 4CO + 6H2O , Delta H(rxn) < 0

We will apply le chateliers principle;

1) O2 increases - reaction will shift product side to reduce the stress at reactant.

2) Deccreasing partial pressure will shift the reaction the side with greater moles i.e. reaction will shift towards product side.

3) Adding 2 moles H2O will shift the reaction at reactant side to reduce the stress at product side.

4) Decrease volume of vessel - reaction will shift the side with fewer moles i.e. reactant side.

5) Catalyst will not affect equilibrium.

6) The reaction is exothermic as Delta H < 0 i.e. negative. Thus increase in temperature will shift the equilibrium towards reactant side.

Let me know if any doubts.

Answer 2

1+1=2 then 2-1=1