Question

Determine delta H at 298K for the reaction C(graphite)+2H2(g)------> CH4(g)

Given: C(graphite)+O2(g) CO2(g) delta H=-393.5KJ/mol
H2+1/2 O2 H2O delta H=-285.8KJ/mol
CO2+2H2OCH4+2O2 delta H=890.3KJ/mol

Answer 1

Given, C(graphite)+O2(g) →CO2(g);    ΔrH°=-393.5kJ mol-1

H2(g) +1/2 O2(g)→H2O(l);    ΔrH°=-285.8 kJ mol-1

CO2(g) +2H2O(l) →CH4(g) +2O2(g);    ΔrH°=890.3 kJ mol-1

Based on the above thermochemical equations, the value of ΔrH° at 298 K for the reaction

C(graphite)+2H2(g)→CH4(g) will be

(a) + 788 kJ mol- 1 (b) + 144.0 kJ mol- 1

(c) - 748 kJ mol- 1 (d) -144.0 kJ mol- 1

Answer 2

Answer:

Explanation:

   

ΔG=ΔH−T Δ S

ΔG=ΔH−T Δ S

                     

0=(170 ×

10³   J)−T(170 J

K

−1

)

0=(170×103 J)−T(170 JK−1)

                    T = 1000 K

                    For spontaneity,  ΔG

ΔG  is  -ve

                    Hence T should be > 1000

therefore T = 1110 K