Question

Consider the production of calcium carbide and liquid water from acetylene:

1/2 Ca(OH)2(s) + 1/2 C2H2(g) 1/2 CaC2(s) + H2O(l)

H = 63.6 kJ

(a) What is the enthalpy change for the reverse reaction?

H_reverse =  kJ/mol.




(b) Balance the forward reaction with whole-number coefficients. What is H for the reaction represented by this equation?

H =  kJ/mol.




(c) Which is more likely to be thermodynamically-favored, the forward reaction or the reverse reaction?

---Select--- The forward reaction. The reverse reaction.




(d) If the (forward) reaction were written to produce water gas, H₂O(g), instead of liquid water, would you expect the magnitude of H to increase (become more positive), decrease (become more negative), or stay the same?

H should  ---Select--- increase decrease stay the same .

Answer 1

a) ∆H for reverse reaction carry opposite sign but same magnitude.

∆Hreverse = -63.6 KJ

b) Balanced equation :

Ca(OH)2(s) + C2H2(g) $\rightleftharpoons$ CaC2(s) + 2H2O(l)

∆H = 2×63.6 = 127.2 KJ

c) Reverse reaction is thermodynamically favoured because it is Exothermic and release energy.

d) If H2O is present in gas form as compared to liquid then ∆H become more positive since energy is needed to vapourize liquid to gas.