Question

Calculate ∆ H of Ethane from the following data .

C + O2 = CO2 ∆H = -393.5 K/J

H2 + 1/2 O2 = H2O ∆ H = -285.8 k/j


C2 H6 + 7/2 O2 = 2CO2 + 3H2O ∆ H = -1560 k/j.


You have to find ∆H of Ethane i,e C2H6.


Pls answer asap!

Answer 1

» Given

C + O2 ---> CO2, ∆H = -393.5 kJ/Mol. ....(1)

H2 + ½O2 ---> H2O, ∆H = -285.8 KJ/Mol. ....(2)

C2H6 + 7/2 O2 ---> 2CO2 + 3H2O, ∆H = -1560 KJ/Mol. ....(3)

» We have to find the ∆H of Ethane, (C2H6)...


Now....

2C + 3H2 ----> C2H6

And we get 2C from eq. (1); 3H2 from eq. (2) and C2H6 from eq. (3)

So,

Eq. (1) + Eq. (2) - Eq.(3)

* Why eq. (3) negative ??

- Because we want the C2H6 on product side and it is given in reactant side...

Now we have to put the values of eq. (1), (2) and (3)

» 2C + 3H2 ---> C2H6

2 (- 393.5) + 3 (- 285.8) - ( -1560)

- 787 - 857.4 + 1560

- 1644.4 + 1560

= - 84.4 KJ/Mol.

=====================================

Answer 2

Given:

• ΔH for the formation of $CO_2$ = -393.5 K/J
• ΔH for the formation of $H_2O$ = -285.8 K/J
• ΔH for the formation of $2CO_2+3H_2O$ = -1560 K/J

To Find:

• ΔH value for the formation of Ethane, $C_2H_6$

Solution:

Let,

$C+O_2 = CO_2$  ⇒ (1)

$H_2+\frac{1}{2} O_2 = H_2O$ ⇒ (2)

$C_2H_6+\frac{7}{2}O_2 = 2CO_2+3H_2O$ ⇒ (3)

Now we need to find the value of ΔH for Ethane.

The reaction for the formation of Ethane is,

$2C+3H_2 = C_2H_6$  

The above reaction is obtained from multiplying equations 1 with 2 and equation 2 with 3 and by adding them and subtracting with equation 3)

ΔH = (2×reaction 1+3×reaction 2-reaction 3)

ΔH = (2×(-393.5)+3×(-285.8)+1560)

ΔH = -787-857.4+1560 = -84.4 K/J

∴ ΔH for the formation of Ethane, $C_2H_6$ = -84.4 K/J.