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A mixture of 0.2 mole of `N_(2)` and 0.6 mole of `H_(2)` react to give `NH_(3)` according to the equation, `N_(2)(g)+3H_(2)(g)iff2NH_(3)(g)` at constant temperature & pressure. 40 % mole of nitrogen react at equilibrium state. Then the ratio of the final volume to the initial volume of gases areA. `4 : 5`B. `5 : 4`C. `7 : 10`D. `8 : 5` |
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Answer» Correct Answer - A `N_(2)(g)+3H_(2)(g)iff2NH_(3)(g)` mole `" 0.2 0.6 "` Left mole `" 0.12 0.36 0.16"` `therefore " "(V_(f))/(V_(i n))=(0.64)/(0.80) " "(because "T and P constant")` `=0.8" or "(4 : 5)` |
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