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For the reaction `N_(2)(g) +3H_(2)(g) to 2NH_(3)(g)` If `Delta[NH_(3)]//Deltat= 4 xx 10^(-8)mol L^(-1)s^(-1)`, what is the value of `Delta[H_(2)]//Deltat`=? |
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Answer» Rate of reaction = `(-d[N_(2)])/(dt) = -1/3 (d[H_(2)])/(dt) = 1/2(d[NH_(3)])/(dt)` `(d[H_(2)])/(dt) = -(3d)/2[NH_(3)]/(dt)` `=-3/2(d[NH_(3)])/(dt) = -3/2 xx (4 xx 10^(-4) mol L^(-1)s^(-1))` `-6 xx 10^(-4)mol L^(-1)s^(-)`. |
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