logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

$H_2O(g) + C(s) \rightarrow CO(g) + H_2(g)$ ; $\Delta H = 131$ kJ. $CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g)$ ; $\Delta H = -282$ kJ. $H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g)$ ; $\Delta H = -242$ kJ. $C(s) + O_2(g) \rightarrow CO_2(g)$ ; $\Delta H = X $ kJ. Based on the above thermochemical equations find the value of X.


(a) +393 kJ
(b) -655 kJ
(c) -393 kJ
(d) +655 kJ

Can you answer this question?
 
 

1 Answer

0 votes
Answer: -393 kJ
$H_2O(g) + C(s) \rightarrow CO(g) + H_2(g)$ ; $\Delta H = 131$ kJ.
$CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g)$ ; $\Delta H = -282$ kJ.
$H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g)$ ; $\Delta H = -242$ kJ.
To obtain this equation, $C(s) + O_2(g) \rightarrow CO_2(g)$, the above equations must be added,
Thus, we need to add the enthalpies of the three equations in order to obtain the enthalpy of the fourth equation.
$\Delta H = X kJ = 131 + (-282) + (-242) = -393$ kJ
$\therefore C(s) + O_2(g) \rightarrow CO_2(g)$ ; $\Delta H = -393 $ kJ.
answered Feb 27, 2014 by mosymeow_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...