返回卡包市场

18 - Rates

暂无描述。系统推荐的高质量记忆内容,适合每天坚持背诵学习。

卡片总数: 24内容版本: v4公开卡包更新时间: 8/1/2026

卡片预览 (24 张)

#1
正面 (问题)

Two ways of determining rate

背面 (解答)

• Measuring the decrease in the concentration of a reactant over time • Measuring the increase in the concentration of a product over time

#2
正面 (问题)

How can rate equations be determines

背面 (解答)

can only be determined experimentally not by mole ratios

#3
正面 (问题)

Rate-concentration = 0 order

背面 (解答)

o Changing the concentration of the chemical has no effect on the rate of the reaction

#4
正面 (问题)

Is the 0 order included in rate equation

背面 (解答)

No

#5
正面 (问题)

Rate from 0 order rate-concentration graph

背面 (解答)

o Rate = K // y-intercept

#6
正面 (问题)

Rate - concentration 1st order

背面 (解答)

o The concentration of the chemical is directly proportional to the rate of reaction, e.g. doubling the concentration of the chemical doubles the rate of reaction

#7
正面 (问题)

Rate from a first order rate-concentration graph

背面 (解答)

o Rate = gradient

#8
正面 (问题)

Rate - concentration 2nd order

背面 (解答)

o The rate is directly proportional to the square of the concentration of that chemical, e.g. doubling the concentration of the chemical increases the rate of reaction by a factor of four

#9
正面 (问题)

Rate from 2nd order rate-concentration graph

背面 (解答)

o Upward curve = plot rate against conc squared = straight line = k = gradient • gradient of straight line

#10
正面 (问题)

Overall order of a rate equation

背面 (解答)

sum of the powers of the reactants in a rate equation

#11
正面 (问题)

Order with respect to [(CH3)3CBr]

卡片正面图片
背面 (解答)

• From the above table, that is experiments 1 and 2 o The [(CH3)3CBr] has doubled, but the [OH-] has remained the same o The rate of the reaction has also doubled o Therefore, the order with respect to [(CH3)3CBr] is 1 (first order)

#12
正面 (问题)

Order with respect to [OH-]

卡片正面图片
背面 (解答)

• From the above table, that is experiments 1 and 3 o The [OH-] has doubled, but the [(CH3)3CBr] has remained the same o The rate of reaction has increased by a factor of 4 (i.e. increased by 22) o Therefore, the order with respect to [OH-] is 2 (second order)

#13
正面 (问题)

Rate equation of this

卡片正面图片
背面 (解答)

Rate = k [(CH3)3CBr] [OH-]2

#14
正面 (问题)

Draw 0 order rate concentration graph

背面 (解答)

卡片背面图片
#15
正面 (问题)

Draw second order rate-concentration graph

背面 (解答)

卡片背面图片
#16
正面 (问题)

Draw third order rate-concentration graph

背面 (解答)

卡片背面图片
#17
正面 (问题)

How to find k

卡片正面图片
背面 (解答)

Use one row from table

卡片背面图片
#18
正面 (问题)

Units of k

背面 (解答)

卡片背面图片
#19
正面 (问题)

Monitoring rate with a colorimeter

背面 (解答)

• In colorimeter – wavelength of light passing through a coloured solution is controlled using a filter • Amount of light absorbed is measures

#20
正面 (问题)

How to use a colorimeter - 6 marks

背面 (解答)

卡片背面图片
#21
正面 (问题)

Concentration - time graph 0 order

背面 (解答)

• In a zero-order reaction, the concentration of the reactant is inversely proportional to time • When the order with respect to a reactant is 0, a change in the concentration of the reactant has no effect on the rate of the reaction

#22
正面 (问题)

Rate constant of concentration time graph

背面 (解答)

gradient= rate constant k

#23
正面 (问题)

Draw 0 order concentration time graph

背面 (解答)

卡片背面图片
#24
正面 (问题)

1st order concentration-time graph

背面 (解答)

• In a first-order reaction, the concentration of the reactant decreases with time o The graph is a curve going downwards and eventually plateaus: