[" The equitibrium constant for the following reaction is "10.5" at "500K" .A system at equilibrium "],[" has "[CO]=0.250M" and "[H_(2)]=0.120M" .What is the "[CH_(3)OH(8)],[[" (a) "0.0378," (b) "0.0435," (c) "0.546]]

The equilibrium constant for the following reaction is 10.5 at 500 K .A syatem at equilibrium has [CO]=0.250M and [H_(2)]=0.120 M "what is the "[CH_(3)OH] ? CO(g)+2H_(2)(g)hArrCH_(3)OH(g)

The equilibrium constant for the following reaction is 10.5 at 500 K .A syatem at equilibrium has [CO]=0.250M and [H_(2)]=0.120 M "what is the "[CH_(3)OH] ? CO(g)+2H_(2)(g)hArrCH_(3)OH(g)

The equilibrium constant for the following reaction is 10.5 at 500 K .A syatem at equilibrium has [CO]=0.250M and [H_(2)]=0.120 M "what is the "[CH_(3)OH] ? CO(g)+2H_(2)(g)hArrCH_(3)OH(g)

The equilibrium constant for the following reactoin is 10 at 500 K.A system at equilibrium has [CO] = 0.25M and [H_(2)] 1M.What is the [CH_(3)OH] ? CO_((g))+2H_(2(g))hArrCH_(3)OH_((g)) CO_((g))+2H_(2(g))hArrCH_(3)OH_(g)

Calculate the equilibrium constant (K) for the formation of NH^ in the following reaction: N_2(g) + 3H_2(g) At equilibrium, the concentration of NH_3 , H_2 and N_2 are 1.2 xx 10^(-2) ,3.0 xx 10^(-2) and 1.5 xx 10^(-2) M respectively.

For the equilibrium system : CO(g)+2H_2(g)hArr CH_3OH(l) what is K_c ?

The following concentrations were obtained for the formation of NH_(3) from N_(2) and H_(2) at equilibrium at 500K . [N_(2)]=1.5xx10^(-2)M . [H_(2)]=3.0xx10^(-2) M and [NH_(3)]=1.2xx10^(-2)M . Calculate equilibrium constant.