For the following three reactions 1, 2 and 3, equilibrium constants are given: (1) CO(g) +H2O(g) leftharpoons CO2(g) +H2(g);K1 (2) CH4(g) +H2O(g) leftharpoons CO(g) +3H2(g);K2 (3) CH4(g) +2H2O(g) leftharpoons CO2(g) +4H2(g);K3 Which of the following relations is correct?

Question

For the following three reactions 1, 2 and 3, equilibrium constants are given: (1) CO(g) +H2O(g) leftharpoons CO2(g) +H2(g);K1 (2) CH4(g) +H2O(g) leftharpoons CO(g) +3H2(g);K2 (3) CH4(g) +2H2O(g) leftharpoons CO2(g) +4H2(g);K3 Which of the following relations is correct? (A) K1√K2=K3 (B) K2=K32 K21 (C) K3 =K1 K2 (D) K3.K32 K21

Tardigrade Admin · Accepted Answer

CO(g) + H2O (g) leftharpoons CO2 (g) +H2 (g) ;K1 . . . . (i) CH4(g) + H2O (g) leftharpoons CO(g) +3H2 (g) ;K2. . . . . (ii) CH4(g) + 2H2O (g) leftharpoons CO2 (g) +4H2 (g) ;K3. . . . . . (iii) Eq. (iii) = eq. (i) + eq. (ii) Thus, K3 = K1 . K2

$K_{1} K_{2} = K_{3}$

$K_{2} = K_{2}^{3} K_{1}^{2}$

$K_{3} = K_{1} K_{2}$

$K_{3} . K_{2}^{3} K_{1}^{2}$

$CO (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g); K_{1} .... (i)$
$C H_{4} (g) + H_{2} O (g) ⇌ CO (g) + 3 H_{2} (g); K_{2} ..... (ii)$
$C H_{4} (g) + 2 H_{2} O (g) ⇌ C O_{2} (g) + 4 H_{2} (g); K_{3} ...... (iii)$
$Eq . (iii) = e q . (i) + e q . (ii)$
Thus, $K_{3} = K_{1} . K_{2}$

Q. For the following three reactions 1, 2 and 3, equilibrium constants are given: (1)CO(g)​+H2​O(g)​⇌CO2(g)​+H2(g)​;K1​ (2)CH4(g)​+H2​O(g)​⇌CO(g)​+3H2(g)​;K2​ (3)CH4(g)​+2H2​O(g)​⇌CO2(g)​+4H2(g)​;K3​ Which of the following relations is correct?

K1​K2​​=K3​

K2​=K23​K12​

K3​=K1​K2​

K3​.K23​K12​

CO(g)+H2​O(g)⇌CO2​(g)+H2​(g);K1​....(i) CH4​(g)+H2​O(g)⇌CO(g)+3H2​(g);K2​.....(ii) CH4​(g)+2H2​O(g)⇌CO2​(g)+4H2​(g);K3​......(iii) Eq.(iii)=eq.(i)+eq.(ii) Thus, K3​=K1​.K2​

$K_{1} K_{2} = K_{3}$

$K_{2} = K_{2}^{3} K_{1}^{2}$

$K_{3} = K_{1} K_{2}$

$K_{3} . K_{2}^{3} K_{1}^{2}$

$CO (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g); K_{1} .... (i)$
$C H_{4} (g) + H_{2} O (g) ⇌ CO (g) + 3 H_{2} (g); K_{2} ..... (ii)$
$C H_{4} (g) + 2 H_{2} O (g) ⇌ C O_{2} (g) + 4 H_{2} (g); K_{3} ...... (iii)$
$Eq . (iii) = e q . (i) + e q . (ii)$
Thus, $K_{3} = K_{1} . K_{2}$