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Question

The endothermic reaction of carbon (graphite) in super heated steam at 500 K
[i.e.,Cgraphite+H2O(g)CO(g)+H2(g)]

is made by providing heat by burning a part of carbon (graphite) in O2(g) followed by an exothermic reaction :
[Cgraphite+O2(g)CO2(g);ΔH=393.7kJmol1at500K].

If a total of 1.34 mole of graphite is required for the production of 1 mole of H2 at 500 K, the heat of reaction for carbon (graphite) with H2O in kJ is ________.

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Solution

One mole of H2 will be formed by 1 mole graphite following the reaction:
C(graphite)+H2O(g)CO(g)+H2(g);ΔH=? ... (i)
Thus, mole of graphite used by the reaction (ii)
=(1.341)=0.34
C(graphite)+O2(g)CO2(g);ΔH=393.7kJ ... (ii)
Therefore,heatproducedin(ii)by0.34moleC(graphite)=heatabsorbedin(i)by1moleC(graphite)
0.34×ΔH(ii)=1×ΔH(i)
or 0.34×393.7=1×ΔH(i)
ΔH(i)=133.86kJ

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