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  • Given C(graphite)+O2(g) → CO2(g) ; rH0=−393.5 kJ mol−1 H2(g)+ <img alt="\frac{1}{2}" src="https://learn.careers360.com/latex-image/?%5Cdpi%7B100%7D%20%5Cfrac%

Given C(graphite)+O2(g) → CO2(g) ; rH0=−393.5 kJ mol−1 H2(g)+ \frac{1}{2} O2(g) → H2O(l) ; rH0=−285.8 kJ mol−1 CO2(g)+2H2O(l) → CH4(g)+2O2(g) ; rH0=+890.3 kJ mol−1 Based on the above thermochemical equations, the value of rH0 at 298 K for the reaction C(graphite)+2H2(g) → CH4(g) will be :
Option: 1  −74.8 kJ mol−1
Option: 2 −144.0 kJ mol−1
Option: 3 +74.8 kJ mol−1
Option: 4 +144.0 kJ mol−1  
 

Answers (1)

best_answer

C+O_{2} \rightarrow CO_{2} , \Delta H = -3.93.5 kJ mol−1

H_{2}+\frac{1}{2}O_{2}\rightarrow H_{2}O , \Delta H =28.5 kJ mol−1

\Delta 2H_{2}+O_{2}\rightarrow 2H_{2}O, \Delta H= 2\times \left ( -285.8 \right ) kJ mol−1

CO_{2}+2H_{2}O \rightarrow CH_{4}+2O_{2}, \Delta H=890.3 kJ mol−1

Adding all equations, we get

C+2H_{2}\rightarrow CH_{4}

\Delta H = 890.3-2\times 285.8-399.6

= -74.8 kJ mol−1

Posted by

vishal kumar

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