logo

Ask Questions, Get Answers

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

Which of the following reactions will define $\Delta H^\circ _f$?


(a) $C(diamond)+O_2(g) \rightarrow CO_2(g)$
(b) $\frac{1}{2}H_2(g) + \frac{1}{2}F_2(g) \rightarrow HF(g)$
(c) $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$
(d) $CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)$

Can you answer this question?
 
 

1 Answer

0 votes
Answer: $\frac{1}{2}H_2(g) + \frac{1}{2}F_2(g) \rightarrow HF(g)$
Standard enthalpy of formation is equal to enthalpy change for the preparation of 1 mol of a compound from its standard state.
Therefore, for the reaction
$\frac{1}{2}H_2(g) + \frac{1}{2}F_2(g) \rightarrow HF(g)$
$\Delta H^\circ _f = \Delta H^\circ$
answered Feb 28, 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
...