| \[C(s)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}CO(g);\] |
| \[\Delta H=-110kJ\] and |
| \[C(s)+{{H}_{2}}O(g)\xrightarrow{{}}CO(g)+{{H}_{2}}(g);\] |
| \[\Delta H=132kJ\] |
| Calculate the mole composition of the mixture of steam and oxygen on being passes over coke at 1273 K, keeping the temperature constant. |
A) 1 : 0.6
B) 0.6 : 1
C) 2 : 3
D) 3 : 2
Correct Answer: A
Solution :
[a] The first reaction is exothermic and the second reaction is endothermic. If on passing the mixture of \[{{O}_{2}}\] and \[{{H}_{2}}O\] (steam) over coke while keeping the temperature constant \[\Delta H\] of both the reactions must be same. Moles of \[{{O}_{2}}\] needed to evolve \[132\text{ }kJ\] \[=\frac{0.5\times 132}{110}=0.6\] Hence steam: \[{{O}_{2}}\] ratio must be 1 : 0.6
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