[A+2B rarr C+D." If "],[[" (18) "-(d[A])/(dt)=5times10^(-4)moll^(-1)s^(-1)1,],[" then "-(d[B])/(dt)" is "]]

A+ 2B rarr C+D." If "-(d[A])/(dt)=5xx10^(-4)" mol L"^(-1) s^(-1)," then "-(d[B])/(dt) is :

A+ 2B rarr C+D." If "-(d[A])/(dt)=5xx10^(-4)" mol L"^(-1) s^(-1)," then "-(d[B])/(dt) is :

In the reaction, A+3Brarr2C,(d[C])/(dt)=2xx10^(-4)"mol.L"^(-1).s^(-1) So , (-d[B])/(dt)("in mol.L"^(-1).s^(-1)) is -

For the reversible reaction A harr B at equilibrium ([B])/([A]) = 6times10^(4) - (d[A])/(dt) = 3times10^(6)s^(-1) [A] - (d[B])/(dt) =K[B] . Thus K is

For the reaction , N_(2)+3H_(2) to 2NH_(3), "if" (d[NH_(3)])/(dt)=2xx10^(-4)"mol"L^(-1)s^(-1), the value of (-d[H_(2)])/(dt) would be

The reaction 3A to 2B+C is carried in a closed vessel. The rate of disappearance of A[(-d[A])/(dt)] is 0.01 molL^(-1)s^(-1) . Calculate (d[B])/(dt) and (d[C])/(dt) .

In the reaction, A + 2B to 6C + 2D , if the initial rate -(d[A])/(dt) at t = 0 is 2.6 xx 10^(-2) m sec^(-1) , what will be the value of -(d[B])/(dt)" at " t = 0 ?

For the reaction in equilibrium A harr B ([B])/([A])=4.0 xx 10^(8)" "(-d[A])/(dt)=2.3 xx 10^(6)S^(-1)[A]" "(-d)/(dt)[B]=K[B] Thus, K is