Question

$K_p$ values for, $C_2{H}_4+H_2\rightleftharpoons C_2{H}_6$, is $5\times {10}^{18}{atm}^{-1}$ and $K_p$ for, $C_2{H}_2+H_2\rightleftharpoons C_2{H}_4$, is $5\times {10}^{26}{atm}^{-1}$ at $50$. If $K_p$ for :
$2H_2+C_2{H}_2\rightleftharpoons C_2{H}_6$ is $x\times {10}^{45}$, then $10x$ is__________.

Answer 1

$C_2{H}_4+H_2\rightleftharpoons C_2{H}_6$, $K_{p1}=5\times {10}^{18}{atm}^{-1}$ ......(1)
$C_2{H}_2+H_2\rightleftharpoons C_2{H}_4$, is $K_{p2}=5\times {10}^{26}{atm}^{-1}$ ......(2)
$2H_2+C_2{H}_2\rightleftharpoons C_2{H}_6$ $K_{p3}=x\times {10}^{45}$ ......(3)
The reaction (3) is obtained by adding reactions (1) and (2).
$K_{p3}=K_{p1}\times K_{p2}=5\times {10}^{18}\times 5\times {10}^{26}=2.5\times {10}^{45}$
Hence, $x=2.5$
$10x=25$