Question

(1) Find the H for the reaction below, given the following reactions and subsequent H values: 2CO2(g) + H2O(g) → C2H2(g) + 5/202(g) CH2(g) + 2H2(g) → CHg) H2O(g) - H2(g) + 1/20, (g) C2H6(g) + 7/202(g) → 2CO2(g) + 3H2O(g) answer - 235 kJ H =-94.5 kJ H =71.2 kJ H --283 kJ (2) Find the H for the reaction below, given the following reactions and subsequent H values: N2H4(I) + H2(g) + 2NH3(g) N2H4(1) + CH4O(1) - CH2O(g) + N2(g) + 3H2(g) N2(g) + 3H2(g) + 2NH3(g) CH401) CH20 g) + H 2(g) answer = -18 kJ H--37 kJ H = -46 kJ H=-65 kJ (3) Find the H for the reaction below, given the following reactions and subsequent H values: H2SO4(l) → SO3(g) + H2O(g) H2S(g) + 2O2(g) → H2SO4(1) H = -235.5 kJ H2S(g) + 2O2(g) → SO3(g) + H2O(1) H=-207 kJ H2O(l) → H2O(g) H = 44 kJ answer = 72 kJ

Answer 1

1) Reversing first equation, sign of H also changes :

C2H6(g) $\rightarrow$ C2H2(g) + 2H2(g) H = 94.5 KJ

Reversing third equation, we get :

2CO2(g) + 3H2O(g) $\rightarrow$ C2H6(g) + 7/2O2(g) H = 283 KJ

Multiplying equation two by 2, and reversing equation we get :

2H2(g) + O2(g) $\rightarrow$ 2H2O(g) H = -2×71.2 = -142.4 KJ

Adding all above equation, we get :

2CO2(g) + H2O(g) $\rightarrow$ C2H2(g) + 5/2O2(g)

H = 94.5+283-142.4 KJ

= 235 KJ