Consider the following reaction mechanism: Step 1: (CH3)3AuPH3 ↔ (CH3)3Au + PH3 (fast) Step 2: (CH3)3Au → C2H6 + (CH3)Au (slow) Step 3: (CH3)Au + PH3 → (CH3)AuPH3 (fast) What is the balanced equation for the overall reaction?

The balanced equation for overall reaction is:

(CH3)3AuPH3 > C2H6 + (CH3)AuPH3

Explanation:

The reaction mechanism is given as follows:

Step 1: (CH3)3AuPH3 ↔ (CH3)3Au + PH3 (fast)

Step 2: (CH3)3Au → C2H6 + (CH3)Au (slow)

Step 3: (CH3)Au + PH3 → (CH3)AuPH3 (fast)

To balance this equation, firstly, we conduct proper atom count for each steps of the reaction mechanism.

It is important to note that for a reaction that involves several steps, the rate law is normally derived from the slow step ( which is step2 from the above mechanism).

Therefore, the balanced chemical equation for the overall reaction is:

(CH3)3AuPH3 > C2H6 + (CH3)AuPH3

(CH3)3AuPH3 > C2H6 + (CH3)AuPH3

Explanation:

Firstly, to balance this equation, we conduct proper atom count for each steps of the reaction mechanism.

Step 1: (CH3)3AuPH3 ↔ (CH3)3Au + PH3 (fast)

Step 2: (CH3)3Au → C2H6 + (CH3)Au (slow)

Step 3: (CH3)Au + PH3 → (CH3)AuPH3 (fast)

It is important to note that for this reaction that involves several steps, the rate- determining step in the rate law is derived from the slow step which is seen in step2 from the above mechanism.

Thus, the balanced chemical equation for the overall reaction is:

(CH3)3AuPH3 > C2H6 + (CH3)AuPH3

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol