Question

1)A well known aphorism in chemistry is that for every 10° increase in temperature, the rate of a reaction will increase twofold. If the reaction were performed at 45° C predict whether it would occur at twice the rate of the reaction performed at 35° C (Show relevant calculations to confirm your answer. If your second set of kinetic trials was performed at a temperature other than 35°, use a temperature exactly 10° higher for the calculation, for example if your second set of trials was run at 38° C, compare 38° to 48°C.

***idk what else is needed to answer this

if youre missing the activation energy its -15.2 kj

Answer 1

We know thatArrhenius equation is expressed as

k = A exp(-E_a / RT)

Now,

T₁ = 35 ^oC = (35 + 273) K = 308 K

T₂ = 45 ^oC = (45 + 273) K = 318 K

So,

K₃₁₈ / k₂₉₅ = [ A exp(-E_a / RT₂) ] / [ A exp(-E_a / RT₁) ]

K₃₁₈ / k₂₉₅ = [ exp(-E_a / RT₂) ] / [ exp(-E_a / RT₁) ]

K₃₁₈ / k₂₉₅ = exp [ (-E_a / RT₂) - exp(-E_a / RT₁) ]

K₃₁₈ / k₂₉₅ = exp [ (E_a / R) { (1/T₁) -(1/T₂) }]

K₃₁₈ / k₂₉₅ = exp [ (-15.2 kJ/ 8.314 J K^-1 mol^-1) { (1/308) -(1/318) }]

K₃₁₈ / k₂₉₅ = exp [ (-15200 J/ 8.314 J K^-1 mol^-1) { (318 -308) /(308 x 318K) }]

K₃₁₈ / k₂₉₅ = exp [ (-1828) { (10) /(308 x 318K) }]

K₃₁₈ / k₂₉₅ = exp (-0.187)

K₃₁₈ / k₂₉₅ = 0.83

Second:

Now,

T₁ = 38 ^oC = (38 + 273) K = 311 K

T₂ = 48 ^oC = (48 + 273) K = 321 K

So,

K₃₂₁ / k₃₁₁ = [ A exp(-E_a / RT₂) ] / [ A exp(-E_a / RT₁) ]

K₃₂₁ / k₃₁₁ = [ exp(-E_a / RT₂) ] / [ exp(-E_a / RT₁) ]

K₃₂₁ / k₃₁₁ = exp [ (-E_a / RT₂) - exp(-E_a / RT₁) ]

K₃₂₁ / k₃₁₁ = exp [ (E_a / R) { (1/T₁) -(1/T₂) }]

K₃₂₁ / k₃₁₁ = exp [ (-15.2 kJ/ 8.314 J K^-1 mol^-1) { (1/311) -(1/321) }]

K₃₂₁ / k₃₁₁ = exp [ (-15200 J/ 8.314 J K^-1 mol^-1) { (321 - 311) /(311 x 321) }]

K₃₂₁ / k₃₁₁ = exp [ (-1828) { (10) /(311 x 321K) }]

K₃₂₁ / k₃₁₁ = exp (-0.183)

K₃₂₁ / k₃₁₁ = 0.83

Note:

The results are unusual, since you provide activation energy its -15.2 kJ (which is negative value).