### JEE Main 12 Jan 2019 Evening Question 15

##### Question: Given:

(i) $ {C _{(graphite)}}+{O _{2(g)}}\to C{O _{2(g)}};{\Delta_r}H^{o}=xkJ,mo{l^{-1}} $

(ii) $ {C _{(graphite)}}+\frac{1}{2}{O _{2(g)}}\to C{O _{(g)}}; $ $ {\Delta_r}H^{o}=ykJmo{l^{-1}} $

(iii) $ C{O _{(g)}}+\frac{1}{2}{O _{2(g)}}\to {CO _2} _{(g)};{\Delta_r}H^{o}=zkJmo{l^{-1}} $

##### Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct? [JEE Main Online Paper Held On 12-Jan-2019 Evening]

#### Options:

A) $ z=x+y $

B) $ x=y-z $

C) $ x=y+z $

D) $ y=2zx $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

- By adding equation (ii) and (iii), we get $ C(graphite)+{O _{2(g)}}\to C{O _{2(g)}}, $ i.e., $ x=y+z $