Question

write the equilibrium constant, Kp for the following gaseous reversible reaction A2(g) +3b2(g) revesible 2Cg​

Answer 1

To find:

equilibrium constant, Kp for the following gaseous reversible reaction:

$\sf{A_{2} + 3B_{2} \: \iff \: 2C_{2}}$

Calculation:

The balanced reaction is:

$\boxed{ \sf{A_{2} + 3B_{2} \: \iff \: 2C_{2}}}$

First , we need to calculate the equilibrium constant for the reaction ;

$\sf{ \therefore \: k_{c} = \dfrac{ { \{ C_{2}\}}^{2} }{ \{A_{2} \} { \{ B_{2}\}}^{3} } }$

Now , let's find k_(p)

$\sf{ \therefore \: k_{p} = k_{c} \times {(RT)}^{\Delta n_{g} } }$

$\sf{ = > \: k_{p} = \dfrac{ { \{ C_{2}\}}^{2} }{ \{A_{2} \} { \{ B_{2}\}}^{3} } \times {(RT)}^{ \{2 - (3 + 1) \}}}$

$\sf{ = > \: k_{p} = \dfrac{ { \{ C_{2}\}}^{2} }{ \{A_{2} \} { \{ B_{2}\}}^{3} } \times {(RT)}^{ \{2 - (4) \}}}$

$\sf{ = > \: k_{p} = \dfrac{ { \{ C_{2}\}}^{2} }{ \{A_{2} \} { \{ B_{2}\}}^{3} } \times {(RT)}^{ - 2}}$

So, final answer is:

$\boxed{ \bf{\: k_{p} = \dfrac{ { \{ C_{2}\}}^{2} }{ \{A_{2} \} { \{ B_{2}\}}^{3} } \times {(RT)}^{ - 2}}}$