Rate of reactions with eggs boiling

[alingy1]

At a high altitude camp in the Rockies, water boils at 95.4°C instead of 100.0°C. A visitor has requested a
soft-boiled egg (usually boiled for 3.00 minutes at 100.0°C). The activation energy for the reaction in question is 453 kJ/mol.
6 marks
egg protein (l) egg protein (s) How long will it take to cook his egg at 95.4°C?

The answers say 18.6min. However, they use this formula:

ln(t2)=ln(t1)+(1/T2-1/T1)x(Ea/R)

I believe the formula should be:
ln(t2)=ln(t1)-(1/T2-1/T1)x(Ea/R)

Am I right?

 

[Chestermiller]

The rates of the reaction at the two temperatures are:

$$r_1=k\exp{\left(-\frac{E}{RT_1}\right)}$$
$$r_2=k\exp{\left(-\frac{E}{RT_2}\right)}$$
We need to have r₁t₁=r₂t₂

So,$$k\exp{\left(-\frac{E}{RT_1}\right)}t_1=k\exp{\left(-\frac{E}{RT_2}\right)}t_2$$
or$$t_2=\exp{\left(\frac{E}{R}(\frac{1}{T_2}-\frac{1}{T_1})\right)}t_1$$
This confirms their formula.

Chet