calculate the change in enthalpy of formation when dihydrogen sulfide gas is bub

Question

calculate the change in enthalpy of formation when dihydrogen sulfide gas is bubbled through a basic solution. V. Raquelle Soto (2014-15) DIRECT METHOD (ENTHALPHY OF REACTION) 1. Calculate the change in enthalpy of formation when dihydrogen sulfide gas is bubbled through a basic solution, given that the heat of formation of each is: HS - 21 kJ/mol, KOH -428.37 kJ/mol, H2O =-285.8 kJ/mol, K2S =-471.5 kJ/mol. The unbalanced reaction is: H2S (g) +2KOH(aq) ,H20 (l) + K2S (aq) 2. Calculate the enthalpy of formation of C2H2 (g), Hr®. given the following information: 2C,H2(g) +502 (g) 4 CO2 (g) + 2H20(1) :-2599 kJ/mol For H2O (1), H°1 For CO2 (g), AH- =-285.8 kJ mor' =-393.5 kJ mor' 91

Explanation / Answer

1.

H2S(g) + 2 KOH(aq) ----------> 2 H2O(l) + K2S(aq)

delta Ho(rkn) = 2 * delta Ho(H2O) + delta Ho(K2S) - delta Ho(H2S) - 2 * delta Ho(KOH)

delta Ho(rkn) = 2 * ( - 285.8) - 471.5 + 21 + 2 * ( 428.37)

delta Ho(rkn) = - 165.36 kJ/mol

2.

2 C2H2(g) + 5 O2(g) -----------> 4 CO2(g) + 2 H2O(l)

delta Ho(rkn) = 4 * delta Ho(CO2) + 2 * delta Ho(H2O) - 2 * delta Ho(C2H2) - 0          ( delta Ho(O2) = 0)

- 2599 = 4 * ( - 393.5 ) + 2 * ( - 285.8) - 2 * delta Ho(C2H2)

- 2599 = - 2145.6 - 2 * delta Ho(C2H2)

2 * delta Ho(C2H2) = ( - 2145.6 + 2599) kJ/mol

2 * delta Ho(C2H2) = 453.4 kJ/mol

delta Ho(C2H2) = 226.7 kJ/mol

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