JEE Main Solved Paper 2017 Question 17
Question: Given $ {C _{(grahite)}}+O _2(g)\to CO _2(g); $ $ {\Delta_r}H^{o}=-393.5,kJ,mo{l^{-1}} $ $ H _2(g)+\frac{1}{2}O _2(g)\to H _2O(l); $ $ {\Delta_r}H^{o}=-285.8,kJ,mo{l^{-1}} $ $ CO _2(g)+2H _2O(l)\to CH _4(g)+2O _2(g); $ $ {\Delta_r}H^{o}=+890.3,kJ,mo{l^{-1}} $ Based on the above thermochemical equations, the value of $ {\Delta_r}H^{o} $ at 298 K for the reaction $ {C _{(grahite)}}+2H _2(g)\to CH _4(g) $ will be:- JEE Main Solved Paper-2017
Options:
A) $ +,74.8,kJ,mo{l^{-1}} $
B) $ +,144.0kJ,mo{l^{-1}} $
C) $ -74.8kJ,mo{l^{-1}} $
D) $ -144.0,kJ,mo{l^{-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ CO _2(g)+2H _2O(\ell )\to CH _4(g)+2O _2(g);{\Delta_r}H^{o}=890.3 $ $ {\Delta_f}H^{o}-393.5-285.8?0 $ $ {\Delta_r}H^{o}={{\sum{({\Delta_f}H^{o})}} _{product}}-{{\sum{({\Delta_f}H^{o})}} _{reactan,ts}} $ $ 890.3=[ 1\times {{({\Delta_f}H^{o})} _{CH _4}}+2\times 0 ]-[1\times (-393.5)+2(-285.8)] $ $ {{({\Delta_f}H^{o})} _{CH _4}}=890.3-965.1=-74.8,kJ/mol $
