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标题: [求助]请问PP-T1-PS-142题 [打印本页]

作者: shining_spags 时间: 2008-6-18 06:58 标题: [求助]请问PP-T1-PS-142题

142.

The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?

(A) 100% decrease

(B) 50% decrease

(D) 40% increase

(E) 50% increase

答案选(D)

能请各位大大解释一下题意及解法吗??实在不能理解,谢谢

作者: cindybaobao 时间: 2008-6-18 20:36

Solution: Let r be the rate of the reaction, x be the concentration of A, and y be the concentration of B. From the given information, we have r = kx^2/y, where k is a constant. Let the original value of x be a and that of y be b. We then have r = ka^2/b.

If b is increased 100%, then the new y is 2b. Substituting it into the equation, we have r = kx^2/2b.

r remain constant. Then ka^2/b = kx^2/2b. This gives x = (sqrt 2)a = 1.4142 . . . a, which shows that a must increase about 40%.

作者: GreenHorse 时间: 2008-6-21 06:48

题目的大概意思是:一个化学反应的速度与物质A的浓度的平方成正比,与物质B的浓度成反比. 现在如果B的物质浓度提高100%(即变成2倍),要使保持此化学反应的速度不变, 那么A物质的浓度应该如何变化.

我是这样思考的:

1)反应的速度与B 的浓度成正比, 当B浓度变成2倍时,若A的浓度不变的时候,反应速度本应变成原来的2倍.

2)按照题目的意思反应速度要不变,必定要让A的浓度的平方也增加2倍.所以A的浓度应该变成原来的(根号2)倍,即A的浓度增加约40%

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