Question 7 Consider the following reaction 8A(g) + 5B(g) = 8C(g) + 6D(g) If C(g)

Question

Question 7

1. Consider the following reaction 8A(g) + 5B(g) = 8C(g) + 6D(g) If C(g) is increasing at the rate of 4.0 mol L the rate of A(g) is [a] mol L,the rate of B(g) is [b] mol L, the rate of D(g) is [d] mol

1. Consider the following reaction 8A(g) + 5B(g) = 8C(g) + 6D(g) If C(g) is increasing at the rate of 4.0 mol L the rate of A(g) is [a] mol L,the rate of B(g) is [b] mol L, the rate of D(g) is [d] mol

Consider the following reaction 8A(g) + 5B(g) = 8C(g) + 6D(g) If C(g) is increasing at the rate of 4.0 mol L the rate of A(g) is [a] mol L,the rate of B(g) is [b] mol L, the rate of D(g) is [d] mol Consider the following reaction 8A(g) + 5B(g) = 8C(g) + 6D(g) If C(g) is increasing at the rate of 4.0 mol L the rate of A(g) is [a] mol L,the rate of B(g) is [b] mol L, the rate of D(g) is [d] mol

Explanation / Answer

(7) 8 A(g) + 5 B(g) => 8 C(g) + 6 D(g)

Rate = (-1/8) x d[A]/dt = (-1/5) x d[B]/dt = (1/8) x d[C]/dt = (1/6) x d[D]/dt

d[C]/dt = 4.0


[a] = d[A]/dt = -d[C]/dt = -4.0 = n4.0


[b] = d[B]/dt = (-5/8) x d[C]/dt = -5/8 x 4.0 = -2.5 = n2.5


[d] = d[D]/dt = (6/8) x d[C]/dt = 6/8 x 4.0 = +3.0 = p3.0


(8) (i) For first order reaction: ln[A] = -kt + ln[A]o

=> Plot of ln[A] vs t gives a straight line with negative slope = -k and y-intercept = ln[A]o

Thus the answer is: (A) First order


(j) For second order reaction: 1/[A] = kt + 1/[A]o

=> Plot of 1/[A] vs t gives a straight line with positive slope = k and y-intercept = 1/[A]o

Thus the answer is: (C) Second order

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.