a) The standard enthalpy change for the following reaction is -270 kJ at 298 K. 2HCN...

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

Solutions

Expert Solution

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

queen_honey_blossom answered 1 year ago

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