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Consider the reaction ` 2NO_((g) ) +O_(2(g) ) to 2NO_(2(g)).` if ` (d [NO_(2)])/(dt)=0.052 M//s ` then `-(d[O_(2)])/(dt)` will beA. `0.052 m//s`B. `0.114m//s`C. `0.026m//s`D. `-0.026 m//s` |
Answer» Correct Answer - C `"Rate"=-(1)/(2)(d[NO])/(dt)=-(d[O_(2)])/(dt)=+(1)/(2) (d[NO_(2)])/(dt)` `-(d[O_(2)])/(dt)=(1)/(2)xx0.052 =0.026 m//s` |
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