Question

question 4 (1 point) saved
for the reaction, xa + yb --> products, the rate law is determined to be r = ka^1b^0, then:
increasing the concentration of b will have no effect on the rate
increasing the concentration of b will increase the rate of the reaction
increasing the concentration of a will have no effect on the rate
the order of the reaction is 0
there is no way to determine the value of k

Explanation:

Step1: Analyze the rate law formula

The rate law is given as \( r = k[A]^1[B]^0 \). Since any non - zero number to the power of \( 0 \) is \( 1 \) (\( x^0=1,x
eq0 \)), the rate law simplifies to \( r = k[A]^1\times1=k[A] \).

Step2: Evaluate the effect of changing [B]

From \( r = k[A] \), we can see that the rate \( r \) does not depend on the concentration of \( B \). Mathematically, if we change \( [B] \), say from \( [B]_1 \) to \( [B]_2 \), since \( [B]^0 = 1 \) for all non - zero \( [B] \), the rate \( r\) remains the same. For example, if \( [A]=a\), \(k = b\), \( [B]_1=c\) and \( [B]_2 = d\) (\(c
eq0,d
eq0\)), \(r_1=b\times a\times c^0=ba\) and \(r_2=b\times a\times d^0=ba\)

Step3: Evaluate the effect of changing [A]

The rate \( r\) is directly proportional to \( [A] \) (\(r = k[A]\)). If we increase \( [A] \) (e.g., from \( [A]_1\) to \( [A]_2\) where \( [A]_2>[A]_1\)), then \(r_2=k[A]_2\) and \(r_1 = k[A]_1\), and \(r_2>r_1\) (since \(k>0\))

Step4: Determine the reaction order

The reaction order is the sum of the exponents in the rate law. Here, the exponent of \( [A]\) is \( 1 \) and the exponent of \( [B]\) is \( 0 \), so the reaction order \(n=1 + 0=1\)

Step5: Consider the rate constant \(k\)

The rate constant \(k\) can be determined experimentally. For example, if we know the rate \(r\) and the concentration of \(A\) (\(r = k[A]\)), then \(k=\frac{r}{[A]}\) (when \( [B]\) is in excess so that its change does not affect the rate measurement in a pseudo - first - order experiment)

Answer:

increasing the concentration of B will have no effect on the rate