In: Chemistry
a) The standard enthalpy change for the following reaction is -270 kJ at 298 K.
2HCN (g) --> 2 C(s, graphite) + H2 (g) + N2 (g) ΔH° = -270 kJ
What is the standard enthalpy change for this reaction at 298 K?
C (s, graphite) + 1/2 H2(g) + 1/2 N2(g) ---> HCN(g)
_______ kJ
2nd reaction can be obtained from 1st reaction by reversing the reaction and multiplying by 1/2
So,
2nd reaction = (-1/2) * 1st reaction
so,
ΔH2° = (-1/2)*ΔH1°
= (-1/2)(-270 kJ)
= 135 KJ
Answer: 135 KJ