Question

For the reaction H2(g) + I2(g) 2HI(g)

• A. $\Delta$ H = + ve
• B. $\Delta$H = – ve
• C. $\Delta$H = zero
• D. $\Delta$H sign cannot be determined

Answer 1

Answer: (B)

$\Delta$H = – ve

Answer 2

at point — A

$\mathrm { Q } = \frac { [ \operatorname { Pr } \text { oduct } ] } { [ \operatorname { Re } \operatorname { ac } \tan t ] } = 0$

So, Q have minimum value at point A.

(II) at point [N2O4] = [NO2] = 0.1 M

$\mathrm { Q } = \frac { \left[ \mathrm { NO } _ { 2 } \right] ^ { 2 } } { \left[ \mathrm {~N} _ { 2 } \mathrm { O } _ { 4 } \right] } = \frac { 0.1 \times 0.1 } { 0.1 } = 0.1$

Q < Kc

H2(g) + I2(g) 2HI(g)

$\log \frac { \mathrm { K } _ { 2 } } { \mathrm {~K} _ { 1 } } = \frac { \Delta \mathrm { H } } { 2.303 } \left[ \frac { 1 } { \mathrm {~T} _ { 1 } } - \frac { 1 } { \mathrm {~T} _ { 2 } } \right]$

$\log \frac { 50 } { 66.9 } = \frac { \Delta \mathrm { H } } { 2.303 } \left[ \frac { 1 } { 623 } - \frac { 1 } { 721 } \right]$

After calculation negative value of $\Delta$ H is obtained.B